(3.1)
Let $E$ denote the event that at least one of the dice lands on 6. So
$$ E=\bset{(6,1),(6,2),(6,3),(6,4),(6,5),(6,6),(1,6),(2,6),(3,6),(4,6),(5,6)} $$
And let $F$ denote the event that the dice land on different numbers:
$$ F=\bset{(i,j):1\leq i,j\leq 6, i\neq j} $$
We can see that $F$ consists of $6\wt6-6=30$ elements. Also the intersection is
$$ EF=\bset{(6,1),(6,2),(6,3),(6,4),(6,5),(1,6),(2,6),(3,6),(4,6),(5,6)} $$
So the intersection $EF$ consists of 10 elements and the conditional probability is
$$ \cp{E}{F}=\frac{\pr{EF}}{\pr{F}}=\frac{10/36}{30/36}=\frac{1}{3} $$
(3.2)
Let $E$ denote the event that the first one of the dice lands on 6. So
$$ E=\bset{(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)} $$
And let $F_i$ denote the event that the sum of the dice is $i$:
$$ F_i=\bset{(j,k):1\leq j,k\leq 6, j+k=i} $$
So
$$ \cp{E}{F_i}=\frac{\pr{EF_i}}{\pr{F_i}} $$
$$ \cp{E}{F_2}=\frac{\pr{EF_2}}{\pr{F_2}}=\frac{\pr{\empty}}{\prb{(1,1)}}=0 $$
$$ \cp{E}{F_3}=\frac{\pr{EF_3}}{\pr{F_3}}=\frac{\pr{\empty}}{\prb{(1,2),(2,1)}}=0 $$
$$ \cp{E}{F_4}=\frac{\pr{EF_4}}{\pr{F_4}}=\frac{\pr{\empty}}{\prb{(1,3),(3,1),(2,2)}}=0 $$
$$ \cp{E}{F_5}=\frac{\pr{EF_5}}{\pr{F_5}}=\frac{\pr{\empty}}{\prb{(1,4),(4,1),(2,3),(3,2)}}=0 $$
$$ \cp{E}{F_6}=\frac{\pr{EF_6}}{\pr{F_6}}=\frac{\pr{\empty}}{\prb{(1,5),(5,1),(2,4),(4,2),(3,3)}}=0 $$
$$ \cp{E}{F_7}=\frac{\pr{EF_7}}{\pr{F_7}}=\frac{\prb{(6,1)}}{\prb{(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)}}=\frac16 $$
$$ \cp{E}{F_8}=\frac{\pr{EF_8}}{\pr{F_8}}=\frac{\prb{(6,2)}}{\prb{(2,6),(6,2),(3,5),(5,3),(4,4)}}=\frac15 $$
$$ \cp{E}{F_9}=\frac{\pr{EF_9}}{\pr{F_9}}=\frac{\prb{(6,3)}}{\prb{(3,6),(6,3),(4,5),(5,4)}}=\frac14 $$
$$ \cp{E}{F_{10}}=\frac{\pr{EF_{10}}}{\pr{F_{10}}}=\frac{\prb{(6,4)}}{\prb{(4,6),(6,4),(5,5)}}=\frac13 $$
$$ \cp{E}{F_{11}}=\frac{\pr{EF_{11}}}{\pr{F_{11}}}=\frac{\prb{(6,5)}}{\prb{(5,6),(6,5)}}=\frac12 $$
$$ \cp{E}{F_{12}}=\frac{\pr{EF_{12}}}{\pr{F_{12}}}=\frac{\prb{(6,6)}}{\prb{(6,6)}}=1 $$
(3.3)
Apparently we are assuming that North & South get dealt first (and together). And then East gets dealt.
Let $E$ denote the event that East has 3 spades and let $F$ denote the event that North and South have a combined total of 8 spades. We wish to compute
$$ \cp{E}{F}=\frac{\pr{EF}}{\pr{F}} $$
Let’s compute $\pr{F}$ first:
$$ \pr{F}=\frac{\binom{13}{8}\binom{39}{18}}{\binom{52}{26}}\approx0.1618 $$
This is just the probability of N&S getting 8 spades and 18 non-spades. And it’s easily computed in python:
from sympy import binomial as binom
binom(13,8)*binom(39,18)/binom(52,26)
Now since we are assuming the N&S get dealt before East, we can count the number of possible hands:
$$ \alpha=\binom{52}{26}\binom{26}{13} $$
And we can count the number of ways for N&S to get 8 spades and East to get 3 spades:
$$ \beta=\binom{13}{8}\binom{39}{18}\binom{5}{3}\binom{21}{10} $$
Hence the probability of N&S getting 8 spades and East getting 3 spades is
$$ \pr{EF}=\frac{\beta}{\alpha}=\frac{\binom{13}{8}\binom{39}{18}\binom{5}{3}\binom{21}{10}}{\binom{52}{26}\binom{26}{13}}\approx0.05488 $$
Hence our desired probability is
$$ \cp{E}{F}=\frac{\pr{EF}}{\pr{F}}\approx\frac{0.05488}{0.1618}\approx0.33918 $$
(3.4)
Let $E$ denote the event that at least one of a pair of fair dice lands on 6. So
$$ E=\bset{(6,1),(6,2),(6,3),(6,4),(6,5),(6,6),(1,6),(2,6),(3,6),(4,6),(5,6)} $$
And let $F_i$ denote the event that the sum of the dice is $i$:
$$ F_i=\bset{(j,k):1\leq j,k\leq 6, j+k=i} $$
So
$$ \cp{E}{F_i}=\frac{\pr{EF_i}}{\pr{F_i}} $$
$$ \cp{E}{F_2}=\frac{\pr{EF_2}}{\pr{F_2}}=\frac{\pr{\empty}}{\prb{(1,1)}}=0 $$
$$ \vdots $$
$$ \cp{E}{F_6}=\frac{\pr{EF_6}}{\pr{F_6}}=\frac{\pr{\empty}}{\prb{(1,5),(5,1),(2,4),(4,2),(3,3)}}=0 $$
$$ \cp{E}{F_7}=\frac{\pr{EF_7}}{\pr{F_7}}=\frac{\prb{(1,6),(6,1)}}{\prb{(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)}}=\frac13 $$
$$ \cp{E}{F_8}=\frac{\pr{EF_8}}{\pr{F_8}}=\frac{\prb{(2,6),(6,2)}}{\prb{(2,6),(6,2),(3,5),(5,3),(4,4)}}=\frac25 $$
$$ \cp{E}{F_9}=\frac{\pr{EF_9}}{\pr{F_9}}=\frac{\prb{(3,6),(6,3)}}{\prb{(3,6),(6,3),(4,5),(5,4)}}=\frac12 $$
$$ \cp{E}{F_{10}}=\frac{\pr{EF_{10}}}{\pr{F_{10}}}=\frac{\prb{(4,6),(6,4)}}{\prb{(4,6),(6,4),(5,5)}}=\frac23 $$
$$ \cp{E}{F_{11}}=\frac{\pr{EF_{11}}}{\pr{F_{11}}}=\frac{\prb{(5,6),(6,5)}}{\prb{(5,6),(6,5)}}=1 $$
$$ \cp{E}{F_{12}}=\frac{\pr{EF_{12}}}{\pr{F_{12}}}=\frac{\prb{(6,6)}}{\prb{(6,6)}}=1 $$
(3.5)
Let $W$ denote the event that the first $2$ balls selected are white. Let $B$ denote the event that the second $2$ balls selected are black. Then the desired probability is $\pr{WB}$. We can compute this as
$$ \pr{WB}=\cp{B}{W}\pr{W} $$
The event $W$ is a hypergeometric random variable:
$$ \pr{W}=\frac{\binom62}{\binom{15}2}=\frac{\frac{6\wts5}{2\wts1}}{\frac{15\wts14}{2\wts1}}=\frac{6\wts5}{15\wts14} $$
The event $B|W$ is also a hypergeometric random variable. But we must first remove $2$ white balls from the population. Hence
$$ \cp{B}{W}=\frac{\binom92}{\binom{13}2}=\frac{\frac{9\wts8}{2\wts1}}{\frac{13\wts12}{2\wts1}}=\frac{9\wts8}{13\wts12} $$
And we get
$$ \pr{WB}=\cp{B}{W}\pr{W}=\frac{6\wts5}{15\wts14}\frac{9\wts8}{13\wts12} $$
Of course we could have simply done this: the probability of selecting the first ball white is $\frac6{15}$, the probability of selecting the second ball white given the first is white is $\frac5{14}$, the probability of selecting the third ball black given the first two are white is $\frac9{13}$, and the probability of selecting the fourth ball black given the first two are white and the third is black is $\frac8{12}$.
In [606]: (winom(9,2)/winom(13,2))*(winom(6,2)/winom(15,2))
Out[606]: 0.0659340659340659
In [607]: 6/15*5/14*9/13*8/12
Out[607]: 0.06593406593406594
(3.6)
Let $E$ denote the event where the first and third drawn balls are white:
$$ E=\bset{(W,b,W,b),(W,W,W,b),(W,b,W,W),(W,W,W,W)} $$
Let $F$ denote the event where exactly 3 of the drawn balls are white:
$$ F=\bset{(b,W,W,W),(W,b,W,W),(W,W,b,W),(W,W,W,b)} $$
Quick solution: notice that every outcome in $F$ and $EF$ is equally likely to occur with or without replacement. Hence the answer in both cases is $\frac24=\frac12$.
With replacement:
And we note that the sample space is of size $16=2^4$ since a black or white ball can be selected at each draw.
$$ \cp{E}{F}=\frac{\pr{EF}}{\pr{F}}=\frac{\prb{(W,b,W,W),(W,W,W,b)}}{4/16}=\frac{2/16}{4/16}=\frac{1}{2} $$
Without replacement:
$$ \cp{E}{F}=\frac{\wN{EF}}{\wN{F}}=\frac24=\frac12 $$
In the case of “without replacement”, nothing changes because of our condition that exactly 3 of the drawn balls are white.
We could have also noticed that the “with replacement” case is a binomial random variable and the “without replacement” case is a hypergeometric random variable.
With replacement:
$$ \pr{F}=\binom43\Bop\frac23\Bcp^3\frac13 $$
For the event $EF$, there are exactly two (disjoint) ways for this two occur: $(1)$ a white ball in position $2$ and a black ball in position $4$ or $(2)$ a black ball in position $2$ and a white ball in position $4$. Both have the probability $\Bop\frac23\Bcp^3\frac13$. Hence
$$ \cp{E}{F}=\frac{\pr{EF}}{\pr{F}}=\frac{2\Bop\frac23\Bcp^3\frac13}{\binom43\Bop\frac23\Bcp^3\frac13}=\frac24=\frac12 $$
In [615]: (2*pw(2/3,3)*1/3)/(winom(4,3)*pw(2/3,3)*1/3)
Out[615]: 0.500000000000000
Without replacement:
$$ \pr{F}=\frac{\binom83\binom41}{\binom{12}4} $$
For the event $EF$, we must again handle the same $2$ disjoint outcomes:
$$ \pr{EF}=\frac8{12}\frac7{11}\frac6{10}\frac49+\frac8{12}\frac4{11}\frac7{10}\frac69 $$
In [616]: (8/12*7/11*6/10*4/9+8/12*4/11*7/10*6/9)/(winom(8,3)*winom(4,1)/winom(12,4))
Out[616]: 0.500000000000000
(3.7)
Let $E$ denote the event that there is exactly one girl in a family of two kids:
$$ \settx{king has $1$ sibling and it's a girl}=E=\bset{(b,g),(g,b)} $$
Let $F$ denote the event that there is at least one boy in a family of two kids:
$$ \settx{king has $1$ sibling}=F=\bset{(b,b),(b,g),(g,b)} $$
$$ \cp{E}{F}=\frac{\pr{EF}}{\prb{(b,b),(b,g),(g,b)}}=\frac{2/4}{3/4}=\frac{2}{3} $$
(3.8)
Let $E$ denote the event that both are girls:
$$ E=\bset{(g,g)} $$
Let $F$ denote the event that the older of the two is a girl:
$$ F=\bset{(g,g),(b,g)} $$
$$ \cp{E}{F}=\frac{\pr{EF}}{\prb{(g,g),(b,g)}}=\frac{1/4}{2/4}=\frac{1}{2} $$
(3.9)
Let $E$ denote the event that the ball chosen from urn $A$ was white:
$$ E=\bset{(w,w,w),(w,w,r),(w,r,w),(w,r,r)} $$
Let $F$ denote the event that exactly 2 white balls were selected:
$$ F=\bset{(w,w,r),(w,r,w),(r,w,w)} $$
Let $W_A$ denote the event that a white ball is selected from urn $A$. Then $W_A^c$ is the event that a red ball is selected from urn $A$.
$$ \cp{E}{F}=\frac{\pr{EF}}{\pr{F}}=\frac{\prb{(w,w,r),(w,r,w)}}{\prb{(w,w,r),(w,r,w),(r,w,w)}} $$
$$ =\frac{\prb{\set{(w,w,r)}\cup\set{(w,r,w)}}}{\prb{\set{(w,w,r)}\cup\set{(w,r,w)}\cup\set{(r,w,w)}}} $$
$$ =\frac{\prb{\set{W_AW_BW_C^c}\cup\set{W_AW_B^cW_C}}}{\prb{\set{W_AW_BW_C^c}\cup\set{W_AW_B^cW_C}\cup\set{W_A^cW_BW_C}}} $$
$$ =\frac{\pr{W_AW_BW_C^c}+\pr{W_AW_B^cW_C}}{\pr{W_AW_BW_C^c}+\pr{W_AW_B^cW_C}+\pr{W_A^cW_BW_C}} \tag{3.9.1} $$
$$ =\frac{\pr{W_A}\pr{W_B}\pr{W_C^c}+\pr{W_A}\pr{W_B^c}\pr{W_C}}{\pr{W_A}\pr{W_B}\pr{W_C^c}+\pr{W_A}\pr{W_B^c}\pr{W_C}+\pr{W_A^c}\pr{W_B}\pr{W_C}} \tag{3.9.2} $$
$$ =\frac{\frac{1}{3}\frac{2}{3}\frac{3}{4}+\frac{1}{3}\frac{1}{3}\frac{1}{4}} {\frac{1}{3}\frac{2}{3}\frac{3}{4}+\frac{1}{3}\frac{1}{3}\frac{1}{4}+\frac{2}{3}\frac{2}{3}\frac{1}{4}}=\frac{\frac6{36}+\frac1{36}}{\frac6{36}+\frac1{36}+\frac4{36}}=\frac{\frac7{36}}{\frac{11}{36}}=\frac7{11} $$
Equation 3.9.1 follows because the events $\set{W_AW_BW_C^c}$, $\set{W_AW_B^cW_C}$, and $\set{W_A^cW_BW_C}$ are pairwise disjoint. Equation 3.9.2 follows because the ball picked from urn $A$ has nothing to do with the ball picked from urn $B$, etc. That is, the three events $W_A$, $W_B$, and $W_C$ are independent.
For posterity’s sake, my original answer:
$$ \cp{E}{F}=\frac{\pr{EF}}{\pr{F}} $$
$$ =\frac{\prb{(w,w,r),(w,r,w)}}{\prb{(w,w,r),(w,r,w),(r,w,w)}} $$
Let’s look at the numerator first:
$$ \prb{(w,r,w),(w,w,r)} $$
$$ =\prt{w from A}\wt\prb{(r,w),(w,r)} \tag{3.9.3} $$
$$ =\prt{w from A}\wt\bop\pr{(r,w)}+\pr{(w,r)}\bcp \tag{3.9.4} $$
$$ =\prt{w from A}\bop\prt{w from B}\prt{r from C}+\prt{r from B}\prt{w from C}\bcp \tag{3.9.5} $$
$$ =\frac{1}{3}\Bop\frac{2}{3}\frac{3}{4}+\frac{1}{3}\frac{1}{4}\Bcp $$
Equation (3.9.3) follows from the independence of picking between different urns. Equation (3.9.4) follows from disjointed events. And (3.9.5) follows from independence. We can compute the denominator similarly to get:
$$ =\frac{\frac{1}{3}\frac{2}{3}\frac{3}{4}+\frac{1}{3}\frac{1}{3}\frac{1}{4}} {\frac{1}{3}\frac{2}{3}\frac{3}{4}+\frac{1}{3}\frac{1}{3}\frac{1}{4}+\frac{2}{3}\frac{2}{3}\frac{1}{4}} $$
$$ =\frac{\frac{1}{6}+\frac{1}{36}} {\frac{1}{6}+\frac{1}{36}+\frac{1}{9}} $$
$$ =0.63636363=\frac{7}{11} $$
(3.10)
I guess this problem is the same as saying “Compute the conditional probability that the THIRD card selected is a spade given that the FIRST and SECOND cards are spades.” That is, I guess this problem is analogous to selecting $3$ cards face down, and then flipping over the second and third cards, without flipping over the first.
There are $52\wt51\wt50$ possible draws, counting permutations. There are $13\wt12\wt50$ ways to choose the cards so that the first and second are spades. And there are $13\wt12\wt11$ ways to select all three spades. So
$$ \cpt{third card a spade}{first and second cards are spades} $$
$$= \frac{\prt{all three cards are spades}}{\prt{first and second cards are spades}} $$
$$= \frac{\frac{13\wt12\wt11}{52\wt51\wt50}}{\frac{50\wt13\wt12}{52\wt51\wt50}} $$
$$= \frac{13\wt12\wt11}{50\wt13\wt12}=\frac{11}{50} $$
Or we could simply see that taking 2 spades out of a deck of 52 leaves 11 spades in a deck of 50.
(3.11)
$$ \wN{B}=4\wts3\dq\wN{A_s}=51+51\dq\wN{BA_s}=3+3 $$
$$ \cp{B}{A_s}=\frac{\pr{BA_s}}{\pr{A_s}}=\frac{\frac6{52\wts51}}{\frac{102}{52\wts51}}=\frac3{51}=\frac1{17} $$
$$ \wN{A}=4\wts48+48\wts4+4\wts3\dq B\subset A\implies\wN{BA}=\wN{B}=4\wts3 $$
$$ \cp{B}{A}=\frac{\pr{BA}}{\pr{A}}=\frac{12}{12(16+16+1)}=\frac1{33} $$
(3.12.a)
Let $E_i$ denote the event that she passes exam $i$ for $i=1,2,3$. The multiplication rule gives us
$$ \prt{she passes all 3 exams}=\pr{E_1E_2E_3} $$
$$ =\pr{E_1}\cp{E_2}{E_1}\cp{E_3}{E_1E_2}=0.9\wts0.8\wts0.7=0.504 $$
(3.12.b)
Note that the event that she failed the second exam is $E_1E_2^c$, and the event that she did not pass all three exams is $(E_1E_2E_3)^c$. Then
$$ \prt{she did not pass all 3}=\pr{(E_1E_2E_3)^c}=1-\pr{E_1E_2E_3}=1-0.504=0.496 $$
Also note that $E_1E_2^c\subset(E_1E_2E_3)^c$: let $x\in E_1E_2^c$. Then $x$ is an outcome where she did not pass all three exams. Hence $x\in(E_1E_2E_3)^c$.
So the desired probability is
$$ \cp{E_1E_2^c}{(E_1E_2E_3)^c}=\frac{\pr{E_1E_2^c(E_1E_2E_3)^c}}{\pr{(E_1E_2E_3)^c}} $$
$$ =\frac{\pr{E_1E_2^c}}{0.496} $$
$$ =\frac{\cp{E_2^c}{E_1}\pr{E_1}}{0.496} $$
$$ =\frac{0.9\wts(1-0.8)}{0.496}=\frac{0.18}{0.496}\approx0.3629 $$
Similarly
$$ \cp{F_1}{G}=\frac{0.1}{0.496}\approx0.20161 $$
$$ \cp{F_3}{G}=\frac{0.9\wts0.8\wts(1-0.7)}{0.496}\approx0.43548 $$
(3.13)
Let $E_i$ be the event that the $i^{th}$ hand has exactly one ace.
$$ \pr{E_1}=\frac{\binom{4}{1}\binom{48}{12}}{\binom{52}{13}} $$
$$ \cp{E_2}{E_1}=\frac{\binom{3}{1}\binom{36}{12}}{\binom{39}{13}} $$
$$ \cp{E_3}{E_1E_2}=\frac{\binom{2}{1}\binom{24}{12}}{\binom{26}{13}} $$
$$ \cp{E_4}{E_1E_2E_3}=\frac{\binom{1}{1}\binom{12}{12}}{\binom{13}{13}}=1 $$
$$ \pr{E_1E_2E_3E_4}=\pr{E_1}\cp{E_2}{E_1}\cp{E_3}{E_1E_2}\cp{E_4}{E_1E_2E_3} $$
$$ =\frac{\binom{4}{1}\binom{48}{12}}{\binom{52}{13}}\frac{\binom{3}{1}\binom{36}{12}}{\binom{39}{13}}\frac{\binom{2}{1}\binom{24}{12}}{\binom{26}{13}}\approx0.105498 $$
(3.14.a)
Let $B_i$, $W_i$ denote the event that a black, white ball is selected on the $i^{th}$ selection.
$$ \pr{B_1}=\frac{\binom71}{\binom{12}1}=\frac{7}{12} $$
$$ \cp{B_2}{B_1}=\frac{9}{14} $$
$$ \cp{W_3}{B_1B_2}=\frac{5}{16} $$
$$ \cp{W_4}{B_1B_2W_3}=\frac{7}{18} $$
$$ \pr{B_1B_2W_3W_4}=\pr{B_1}\cp{B_2}{B_1}\cp{W_3}{B_1B_2}\cp{W_4}{B_1B_2W_3} $$
$$ =\frac{7}{12}\frac{9}{14}\frac{5}{16}\frac{7}{18}\approx0.0455729 \tag{3.14.a.1} $$
(3.14.b)
We can simply note that there are $\binom42$ combinations of exactly $2$ black balls and for any such combination, the numerators and denominators in equation (3.14.a.1) stay the same. Hence the answer is $\binom42\frac{7}{12}\frac{9}{14}\frac{5}{16}\frac{7}{18}$.
Or we can compute this more explicitly:
$$ \prt{of the first 4 selected balls, exactly two are black} $$
$$ =\prt{balls 1,2 black}+\prt{balls 1,3 black}+\prt{balls 1,4 black} $$
$$ +\prt{balls 2,3 black}+\prt{balls 2,4 black}+\prt{balls 3,4 black} $$
$$ =\pr{B_1B_2W_3W_4}+\pr{B_1W_2B_3W_4}+\pr{B_1W_2W_3B_4} $$
$$ +\pr{W_1B_2B_3W_4}+\pr{W_1B_2W_3B_4}+\pr{W_1W_2B_3B_4} $$
$$ =\frac{7}{12}\frac{9}{14}\frac{5}{16}\frac{7}{18}+\frac{7}{12}\frac{5}{14}\frac{9}{16}\frac{7}{18}+ \dots+\frac{5}{12}\frac{7}{14}\frac{7}{16}\frac{9}{18} $$
$$ \approx6\wts0.0455729=0.2734374 $$
(3.15)
The wording to this question is very tricky. The is the question: given that a randomly selected woman has an ectopic pregnany, what is the probability that she’s a smoker? Or, what is the probability that she’s a smoker, given that she had an ectopic pregnancy?
Let $E$ be the event that a randomly chosen pregnant women has an ectopic pregnancy and $S$ the event that the chosen person is a smoker. Then the problem states that
$$ \cp{E}{S}=2\cp{E}{S^c}\dq\pr{S}=0.32 $$
So the conditional probability is
$$ \cp{S}{E}=\frac{\pr{SE}}{\pr{E}} $$
$$ =\frac{\cp{E}{S}\pr{S}}{\cp{E}{S}\pr{S}+\cp{E}{S^c}\pr{S^c}} \tag{Bayes Formula} $$
$$ =\frac{2\cp{E}{S^c}\pr{S}}{2\cp{E}{S^c}\pr{S}+\cp{E}{S^c}\pr{S^c}} $$
$$ =\frac{2\cp{E}{S^c}\pr{S}}{\cp{E}{S^c}\bop2\pr{S}+\pr{S^c}\bcp} $$
$$ =\frac{2\pr{S}}{2\pr{S}+\pr{S^c}} $$
$$ =\frac{0.64}{0.64+(1-0.32)}=\frac{0.64}{0.64+0.68}=\frac{0.64}{1.32}\approx0.48484 $$
(3.16)
Let $S$ be the event that the baby of a randomly chosen pregnant women survives delivery and $C$ the event that the chosen pregnant woman has a C-section. Then the problem states that
$$ \cp{S}{C}=0.96\dq \pr{S}=0.98\dq \pr{C}=0.15 $$
So
$$ 0.98=\pr{S}=\cp{S}{C}\pr{C}+\cp{S}{C^c}\pr{C^c} $$
$$ =0.96\wts0.15+\cp{S}{C^c}0.85 $$
$\iff$
$$ \cp{S}{C^c}0.85=0.98-0.144=0.836 $$
$\iff$
$$ \cp{S}{C^c}=\frac{0.836}{0.85}\approx0.98353 $$
(3.17.a)
Let $D$ denote the event that a random family owns a dog and let $C$ denote the event that a random family owns a cat. We’re given:
$$ \pr{D}=0.36\dq\cp{C}{D}=0.22\dq\pr{C}=0.3 $$
$$ \pr{DC}=\cp{C}{D}\wts\pr{D}=0.22\wts0.36=0.0792 $$
(3.17.b)
$$ \cp{D}{C}=\frac{\pr{DC}}{\pr{C}}=\frac{0.0792}{0.3}=0.264 $$
(3.18)
Given:
$$ \pr{I}=0.46\dq\pr{L}=0.3\dq\pr{C}=0.24 $$
Notice that $I$, $L$, and $C$ are mutually exclusive and exhaustive. Also given:
$$ \cp{V}{I}=0.35\dq\cp{V}{L}=0.62\dq\cp{V}{C}=0.58 $$
So we can compute $\pr{V}$:
$$ \pr{V}=\cp{V}{I}\pr{I}+\cp{V}{L}\pr{L}+\cp{V}{C}\pr{C} $$
$$ =0.35\wts0.46+0.62\wts0.3+0.58\wts0.24=0.4862 $$
And easy Bayes Formula computations for each conditional probability:
$$ \cp{I}{V}=\frac{\pr{IV}}{\pr{V}}=\frac{\cp{V}{I}\pr{I}}{0.4862}\approx0.3311 $$
$$ \cp{L}{V}=\frac{\cp{V}{L}\pr{L}}{0.4862}\approx0.3826 $$
$$ \cp{C}{V}=\frac{\cp{V}{C}\pr{C}}{0.4862}\approx0.2863 $$
(3.19)
Given:
$$ \cp{A}{W}=0.48\dq\cp{A}{M}=0.37\dq\pr{W}=0.38\dq\pr{M}=0.62 $$
Probability of attending party:
$$ \pr{A}=\cp{A}{W}\pr{W}+\cp{A}{M}\pr{M}=0.1824+0.2294=0.4118 $$
Conditional probability that random chosen person at party is a woman:
$$ \cp{W}{A}=\frac{\cp{A}{W}\pr{W}}{\pr{A}}=\frac{0.1824}{0.4118}\approx0.443 $$
(3.20)
Given:
$$ \pr{F}=0.52\dq\pr{C}=0.05\dq\pr{FC}=0.02 $$
Compute:
$$ \cp{F}{C}=\frac{0.02}{0.05}=0.4\dq\cp{C}{F}=\frac{0.02}{0.52}=0.0385 $$
Extra:
$$ \cp{M}{C}=\cp{F^c}{C}=1-0.4=0.6 $$
$$ \cp{C}{M}=\frac{\pr{MC}}{\pr{M}}=\frac{\cp{M}{C}\pr{C}}{1-\pr{F}}=\frac{0.6\wts0.05}{0.48}=0.0625 $$
Check:
$$ \cp{C}{F}\pr{F}+\cp{C}{M}\pr{M}=0.0385\wts0.52+0.0625\wts0.48=0.05=\pr{C} $$
(3.21)
Given:
$$ \pr{W_LH_L}=\frac{212}{500}=0.424 \quad\quad \pr{W_LH_M}=\frac{198}{500}=0.396 $$
$$ \pr{W_MH_L}=\frac{36}{500}=0.072 \quad\quad \pr{W_MH_M}=\frac{54}{500}=0.108 $$
Compute:
$$ \pr{H_L}=\pr{W_LH_L}+\pr{W_MH_L}=0.424+0.072=0.496 $$
$$ \pr{H_M}=1-\pr{H_L}=0.504 $$
$$ \cp{W_M}{H_M}=\frac{\pr{W_MH_M}}{\pr{H_M}}=\frac{0.108}{0.504}\approx0.2143 $$
$$ \cp{W_M}{H_L}=\frac{\pr{W_MH_L}}{\pr{H_L}}=\frac{0.072}{0.496}\approx0.1452 $$
(3.22.a)
Let $B$ be the number that the blue die lands on, let $Y$ be the number that the yellow die lands on, and let $R$ be the number that the red die lands on. And let $\Psi$ denote the event that all dice land on different values.
$$ \pr{\Psi}=\frac{6\wt5\wt4}{6\wt6\wt6}=\frac{120}{216}=\frac{5}{9} $$
Nice sanity check:
$$ \binom306\wts5\wts4+\binom326\wts5+\binom336=120+90+6=216 $$
In words:
You can get all $3$ different (abc - $1$ combination of the dice) in $6\wt5\wt4=120$ ways.
You can get the first two equal and the third different (aab) in $6\wt1\wt5=30$ ways. Similarly (aba) can occur in $6\wt5\wt1=30$ ways and (baa) can occur in $6\wt5\wt1=30$ ways. Since these are mutually exclusive events, the total count of two equal dice and one different is $30+30+30=90$.
You can get the same value (aaa - $1$ combination of the dice) on all $3$ dice in $6\wt1\wt1=6$ ways.
Sanity check of mutually exlusive and exhaustive events: $120+90+6=216$.
(3.22.b)
First notice that $\set{B<Y<R}\subset\Psi$. So
$$ \cp{B<Y<R}{\Psi}=\frac{\pr{B<Y<R}}{\pr{\Psi}} $$
We can count out the event $\set{B<Y<R}$:
- $123$
- $124$
- $125$
- $126$
- $134$
- $135$
- $136$
- $145$
- $146$
- $156$
- $234$
- $235$
- $236$
- $245$
- $246$
- $256$
- $345$
- $346$
- $356$
- $456$
In hindsight, we see the count is simply $\binom63$ since we merely counted the combinations of choosing $3$ numbers from $1,2,3,4,5,6$. Hence
$$ \cp{B<Y<R}{\Psi}=\frac{20}{6\wts5\wts4}=\frac16 $$
Alternative approach: In the event that no two dice are equal, there are $3!=6$ possible permutations:
- $B<Y<R$
- $B<R<Y$
- $Y<B<R$
- $Y<R<B$
- $R<Y<B$
- $R<B<Y$
Each permutation contains the same number rolls. Hence \(\cp{B<Y<R}{\Psi}=\frac{1}{6}\)
(3.22.c)
$$ \pr{B<Y<R}=\cp{B<Y<R}{\Psi}\pr{\Psi}=\frac16\frac59=\frac5{54} $$
Alternatively:
$$ \pr{B<Y<R}=\frac{\wN{B<Y<R}}{\wN{\Omega}}=\frac{20}{6^3}=\frac{5\wts4}{6\wts9\wts4}=\frac5{54} $$
(3.23.a)
$$ \pr{2_w}=\cp{2_w}{1_w}\pr{1_w}+\cp{2_w}{1_w^c}\pr{1_w^c}=\frac{2}{3}\frac{1}{3}+\frac{1}{3}\frac{2}{3}=\frac{4}{9} $$
(3.23.b)
$$ \cp{1_w}{2_w}=\frac{\pr{1_w2_w}}{\pr{2_w}}=\frac{\cp{2_w}{1_w}\pr{1_w}}{\frac{4}{9}}=\frac{2}{9}\frac{9}{4}=\frac{1}{2} $$
(3.24.a)
Note that each outcome in the sample space $S$ is equally likely:
$$ S=\bset{(bg),(bb),(gb),(gg)} $$
Let $E$ denote the event that at least one of the balls is painted gold:
$$ E=\bset{(bg),(gb),(gg)} $$
Let $G$ denote the event that both balls are painted gold:
$$ G=\bset{(gg)} $$
$$ \cp{G}{E}=\frac{\pr{GE}}{\pr{E}}=\frac{\pr{G}}{\frac{3}{4}}=\frac{1}{3} $$
(3.24.b)
The solutions manual gives: “Since we have no information about the ball in the urn, the answer is $\frac12$.” I guess he’s implying that we can indentify the ball number, $1$ or $2$, of the ball that fell out? And hence the other ball is $50-50$.
So I guess the correct answer depends. If we can tell that the ball that falls out is the first ball, then we get the following:
Let $G_1$ denote the event that the first ball is painted gold:
$$ G_1=\bset{(gb),(gg)} $$
$$ \cp{G}{G_1}=\frac{\pr{GG_1}}{\pr{G_1}}=\frac{\pr{G}}{\frac{2}{4}}=\frac{1}{2} $$
And similar for $G_2$: $\cp{G}{G_2}=\frac{1}{2}$.
But if we cannot tell the difference between balls 1 and 2, then the answer is the same as in part (a): we found out that at least one of the balls is gold so the answer is $\frac13$.
(3.26)
Bayes Formula:
$$ \cp{M}{C}=\frac{\cp{C}{M}\pr{M}}{\cp{C}{M}\pr{M}+\cp{C}{F}\pr{F}}=\frac{0.05\wts0.5}{0.05\wts0.5+0.0025\wts0.5} $$
In [640]: (0.05*0.5)/(0.05*0.5+0.0025*0.5)
Out[640]: 0.9523809523809523
In general, if $m=\pr{M}$ then
$$ \cp{M}{C}=\frac{0.05m}{0.05m+0.0025(1-m)} $$
So if the population consisted of twice as many males as females:
$$ m=2(1-m)=2-2m\iff3m=2\iff m=\frac23 $$
and
In [641]: m=2/3
In [642]: (0.05*m)/(0.05*m+0.0025*(1-m))
Out[642]: 0.9756097560975611
(3.28.a)
Let $A$ denote the event that the $20^{th}$ flip gives the first ace. Let $S$ denote the event that the first ace is the ace of spades. Let $F$ denote the event that the ace of spades is the first flip after the first ace. Conditioning on the conditional probability (p.94), we have
$$ \cp{F}{A}=\cp{F}{SA}\cp{S}{A}+\cp{F}{S^cA}\cp{S^c}{A} $$
$SA$ denotes the event the first ace appears on the $20^{th}$ flip and it’s the ace of spades. Given this, there is no chance of the ace of spades being the first flip after the first ace. Hence $\cp{F}{SA}=0$.
$\cp{S^c}{A}$ is the probability that the first ace, given it appears on the $20^{th}$ flip, is the ace of clubs, diamonds, or hearts. So this is just $\frac34$.
$S^cA$ is the event that the first ace is drawn on the $20^{th}$ flip and it’s the ace of clubs, diamonds, or hearts. So $\cp{F}{S^cA}$ is the probability that the ace of spades is the $21^{st}$ card drawn given that it wasn’t drawn on the first $20$ flips. With $32$ cards remaining and the ace of spades being one of them, there is a probability of $\frac1{32}$ of selecting the ace of spades. Hence
$$ \cp{F}{A}=0+\cp{F}{S^cA}\cp{S^c}{A}=\frac{1}{32}\frac{3}{4}=\frac{3}{128}\approx0.0234 $$
(3.28.b)
Here we use the shorter formula for conditioning on a conditional probability:
$$ Q(E_1) = Q(E_1|E_2)Q(E_2) + Q(E_1|E_2^c)Q(E_2^c) $$
Throughout, it’s given that the first ace was picked $20^{th}$. Let $N$ denote the event that the next card picked is the 2 of clubs. And let $B$ be the event that the two of clubs appeared before the first ace.
$$ \pr{N}=\cp{N}{B}\pr{B}+\cp{N}{B^c}\pr{B^c}=0\wts\frac{19}{48}+\frac{1}{32}\frac{29}{48} $$
$\cp{N}{B}=0$ because the 2 of clubs cannot be drawn next if it was drawn before. $\pr{B}=\frac{19}{48}$ because it is given that an ace was selected $20^{th}$ (hence $19$) and no aces were selected in the first $19$ (hence $48$).
$\cp{N}{B^c}=\frac{1}{32}$ because there are $32$ cards remaining, of which the 2 of clubs is one. And $\pr{B^c}=1-\pr{B}=1-\frac{19}{48}=\frac{29}{48}$. Hence
$$ \pr{N}=\frac{1}{32}\frac{29}{48}=\frac{29}{1536}\approx0.0189 $$
(3.29)
Let $A$ denote the event that none of the final 3 balls chosen were ever used. So we want to compute $\pr{A}$.
And let $B_i$ denote the event that $i$ of the first $3$ balls chosen had previously been used. That is, $B_0$ means none of the first $3$ balls were previously used (i.e. all were new). $B_1$ means $1$ of the first $3$ balls was previously used (i.e. two were new). $B_2$ means $2$ of the first $3$ balls were previously used (i.e. one was new). $B_3$ means all $3$ of the first $3$ balls were previously used (i.e. none were new).
The $B_i$ are mutually exclusive and exhaustive (in the sample space of the first $3$ chosen balls). So we can condition $A$ on the $B_i$:
$$ \pr{A}=\condi{A}{B_0}+\condi{A}{B_1}+\condi{A}{B_2}+\condi{A}{B_3} \tag{3.29.1} $$
So let’s compute values for each of these terms:
$$ \pr{B_0}=\frac{\binom{6}{0}\binom{9}{3}}{\binom{15}{3}} $$
This follows because we want to select $0$ of the $6$ original used balls. And we want to select $3$ of the $9$ original new balls. Now given than none of the first three chosen balls was previously used, once chosen, these balls are now used and returned to the box. So now the box has $9$ used balls and $6$ new balls. So the probably of choosing 3 new balls is $\binom{6}{3}\big/\binom{15}{3}$. That is
$$ \cp{A}{B_0}=\frac{\binom{6}{3}}{\binom{15}{3}} $$
Similarly we get
$$ \pr{B_1}=\frac{\binom{6}{1}\binom{9}{2}}{\binom{15}{3}} $$
since $B_1$ means we select $1$ used and $2$ new balls from the original box. Now we return those balls to the box, all used, so that the box now contains $8$ used and $7$ new. So the probability of selecting $3$ new balls given that $2$ of the original, new balls were converted to used balls is
$$ \cp{A}{B_1}=\frac{\binom{7}{3}}{\binom{15}{3}} $$
Similarly we get
$$ \pr{B_2}=\frac{\binom{6}{2}\binom{9}{1}}{\binom{15}{3}} \dq \cp{A}{B_2}=\frac{\binom{8}{3}}{\binom{15}{3}} $$
and
$$ \pr{B_3}=\frac{\binom{6}{3}\binom{9}{0}}{\binom{15}{3}} \dq \cp{A}{B_3}=\frac{\binom{9}{3}}{\binom{15}{3}} $$
We see that this can be generalized:
$$ \pr{B_i}=\frac{\binom{6}{i}\binom{9}{3-i}}{\binom{15}{3}} \quad\quad \cp{A}{B_i}=\frac{\binom{6+i}{3}}{\binom{15}{3}} $$
So that (3.29.1) becomes
$$ \pr{A}=\sum_{i=0}^{3}\pr{B_i}\cp{A}{B_i} $$
$$ =\sum_{i=0}^{3} \frac{\binom{6}{i}\binom{9}{3-i}}{\binom{15}{3}} \frac{\binom{6+i}{3}}{\binom{15}{3}} $$
$$ =\frac{1}{\bop\binom{15}{3}\bcp^2}\sum_{i=0}^{3} \binom{6}{i}\binom{9}{3-i} \binom{6+i}{3} $$
$$ =\frac{1}{455^2}\sum_{i=0}^{3} \binom{6}{i}\binom{9}{3-i} \binom{6+i}{3} $$
$$ =0.08926 $$
This is easily computed in python:
In [1050]: from sympy import binomial as binom
In [1051]: import numpy as np
In [1052]: def tbb():
...: c=1/(np.power(455,2))
...: r=0
...: for i in range(4):
...: pr=binom(6,i)*binom(9,3-i)*binom(6+i,3)
...: r+=pr
...: return r*c
...:
In [1053]: tbb()
Out[1053]: 0.0892645815722739
In [1054]: # or
In [1055]: def tbb2():
...: c=1/(np.power(455,2))
...: r=0
...: for i in range(4):
...: pr=binom(6,3-i)*binom(9,i)*binom(9-i,3)
...: r+=pr
...: return r*c
...:
In [1056]: tbb2()
Out[1056]: 0.0892645815722739
(3.30)
$$ \pr{B}=\cond{B}{1}=\frac12\frac12+\frac23\frac12=\frac14+\frac13=\frac3{12}+\frac4{12}=\frac7{12} $$
$$ \cp{1}{B^c}=\frac{\pr{1B^c}}{\pr{B^c}}=\frac{\cp{B^c}{1}\pr{1}}{1-\pr{B}}=\frac{\frac12\frac12}{\frac5{12}}=\frac14\frac{12}5=\frac35 $$
(3.31)
Let $C$ be the event that the tumor is cancerous, and let $N$ be the event that the doctor does not call. Then
$$ \alpha=\pr{C} $$
$$ \beta=\cp{C}{N} $$
$\beta$ should be larger than $\alpha$ because if the doctor had called, she would know that she doesn’t have cancer. But the doctor didn’t call ($\beta$) so this effectively reduces to the sample space while not changing the underlying likelihood that the tumor is cancerous. Said another way: the no-call implies decreased chance of good news.
$$ \alpha=\pr{C}=\cond{C}{N} $$
But $\cp{C}{N^c}=0$ because the doctor calling must mean good news. So
$$ \alpha=\cp{C}{N}\pr{N}=\beta\pr{N}\leq\beta $$
Alternative approach: Bayes Formula: Let $H$ denote the event that the coin flip gives heads:
$$ \beta=\cp{C}{N}=\frac{\cp{N}{C}\pr{C}}{\cp{N}{C}\pr{C}+\cp{N}{C^c}\pr{C^c}}=\frac{1\alpha}{1\alpha+\pr{H^c}(1-\alpha)} $$
$$ =\frac{\alpha}{\alpha+\frac12(1-\alpha)}=\frac\alpha{\alpha+\frac12-\frac12\alpha}=\frac\alpha{\frac12\alpha+\frac12}=\frac{2\alpha}{\alpha+1} $$
The same inequality $\alpha\leq\beta$ holds. When $\alpha=0$, this clearly holds. When $\alpha>0$:
$$ \alpha\leq\frac{2\alpha}{\alpha+1}\iff\alpha(\alpha+1)\leq2\alpha\iff\alpha+1\leq2\iff\alpha\leq1 $$
Yet another alternative approach: Condition on the conditional probability.
$$ \beta=\cp{C}{N}=\cp{C}{HN}\cp{H}{N}+\cp{C}{H^cN}\cp{H^c}{N} $$
$HN$ denotes the event that heads is flipped and no call is made. But when heads occurs, the doctor doesn’t make a call only if the patient has cancer. That is, the occurrence of $HN$ implies the occurrence of $C$. Hence $\cp{C}{HN}=1$.
$H^cN$ denotes the event that tails is flipped and no call is made. But when tails occurs, the doctor doesn’t make a call regardless of whether or not the patient has cancer. That is, the occurrence of $H^cN$ gives us no information that will help determine the probability that $C$ occurs. Hence $\cp{C}{H^cN}=\pr{C}=\alpha$.
We can use Bayes Formula next:
$$ \cp{H}{N}=\frac{\cp{N}{H}\pr{H}}{\cp{N}{H}\pr{H}+\cp{N}{H^c}\pr{H^c}}=\frac{\alpha\frac12}{\alpha\frac12+1\wts\frac12}=\frac{\alpha}{\alpha+1} $$
And
$$ \cp{H^c}{N}=1-\frac{\alpha}{\alpha+1}=\frac{\alpha+1-\alpha}{\alpha+1}=\frac1{\alpha+1} $$
Putting it all together:
$$ \beta=\cp{C}{N}=1\frac{\alpha}{\alpha+1}+\alpha\frac1{\alpha+1}=\frac{2\alpha}{\alpha+1} $$
(3.32)
Let $E$ denote the event that the chosen child is the eldest and let $N_i$ denote the event that the family of the chosen child has $i$ children. Bayes Formula gives us
$$ \cp{N_i}{E}=\frac{\cp{E}{N_i}\pr{N_i}}{\sum_{j=1}^{4}\condi{E}{N_j}}=\frac{\frac1i\wts p_i}{\sum_{j=1}^{4}\frac{p_j}j} $$
In [700]: p=[0,.1,.25,.35,.3]
In [701]: nec=lambda i=1: (p[i]/i)/sum([p[j]/j for j in range(1,5)])
In [702]: nec(),nec(4)
Out[702]: (0.24, 0.18)
Since a chosen child is equally likely to be the youngest or oldest (or any in between, when applicable), then
$$ \cp{Y}{N_i}=\cp{E}{N_i}=\frac1i $$
and the computations and resulting probabilities are the same.
(3.33.a)
Note that
$$ \cp{E}{R}=1-\cp{E^c}{R}=1-0.3=0.7 $$
$$ \cp{E}{R^c}=1-\cp{E^c}{R^c}=1-0.1=0.9 $$
So that
$$ \pr{E}=\cond{E}{R}=0.7\wts0.7+0.9\wts0.3=0.76 $$
(3.33.b)
$$ \cp{R}{E}=\frac{\pr{RE}}{\pr{E}}=\frac{\cp{E}{R}\pr{R}}{0.76}=\frac{0.49}{0.76}\approx0.6447 $$
(3.34)
Let $G$ denote the event that the suspect is guilty and let $C$ denote the event that he possesses the characteristic of the criminal.
Previously, when it was certain that the criminal possessed the characteristic, we computed $\cp{C}{G}=1$. This says that, given the suspect is guilty, then he must possess the characteristic, since the criminal possesses the characteristic.
Now we see that there is a $90\%$ likelihood that the criminal possesses the characteristic. So, given that the suspect is guilty, then it must be that he possesses the characteristic with $90\%$ likelihood.
$$ \cp{G}{C}=\frac{\cp{C}{G}\pr{G}}{\cp{C}{G}\pr{G}+\cp{C}{G^c}\pr{G^c}} $$
$$ =\frac{0.9\wts0.6}{0.9\wts0.6+0.2\wts0.4}=\frac{\frac{54}{100}}{\frac{54}{100}+\frac8{100}}=\frac{54}{62} $$
(3.35.a)
$$ \pr{U}=\condi{U}{M}+\condi{U}{D}=0.7\wts0.6+0.5\wts0.4 $$
$$ =\frac{42}{100}+\frac{20}{100}=\frac{62}{100}=\frac{31}{50} $$
(3.35.b)
$$ \cp{D}{U^c}=\frac{\condi{U^c}{D}}{\condi{U^c}{D}+\condi{U^c}{M}} $$
$$ =\frac{0.5\wts0.4}{0.5\wts0.4+0.3\wts0.6}=\frac{0.2}{0.2+0.18}=\frac{0.2}{0.38}=\frac{20}{38}=\frac{10}{19} $$
(3.36)
Each store has the following number of women working there:
- $0.5\wts50=25$
- $0.6\wts75=45$
- $0.7\wts100=70$
The probability that she works in store $C$ is $\frac{70}{140}=\frac12$.
Alternatively, we can use Bayes Formula:
$$ \cpttwo{C}{woman}=\frac{\cptone{woman}{C}\pr{C}}{\cptone{woman}{A}\pr{A}+\cptone{woman}{B}\pr{B}+\cptone{woman}{C}\pr{C}} $$
$$ =\frac{0.7\wts\frac{100}{225}}{0.5\wts\frac{50}{225}+0.6\wts\frac{75}{225}+0.7\wts\frac{100}{225}}=\frac{70}{25+45+70}=\frac12 $$
(3.37.a)
$$ \cptone{fair}{h}=\frac{\cpttwo{h}{fair}\prt{fair}}{\cpttwo{h}{fair}\prt{fair}+\cpttwo{h}{unfair}\prt{unfair}} $$
$$ =\frac{\frac12\frac12}{\frac12\frac12+1\wts\frac12}=\frac{\frac14}{\frac34}=\frac13 $$
(3.37.b)
$$ \cptone{fair}{h,h}=\frac{\cpttwo{h,h}{fair}\prt{fair}}{\cpttwo{h,h}{fair}\prt{fair}+\cpttwo{h,h}{unfair}\prt{unfair}} $$
$$ =\frac{\cpttwo{h}{fair}\cpttwo{h}{fair}\prt{fair}}{\cpttwo{h}{fair}\cpttwo{h}{fair}\prt{fair}+\cpttwo{h}{unfair}\cpttwo{h}{unfair}\prt{unfair}} $$
$$ =\frac{\frac12\frac12\frac12}{\frac12\frac12\frac12+1\wts1\wts\frac12}=\frac{\frac18}{\frac58}=\frac15 $$
Notice in the second equality we used independence and the fact that conditional probabilities are probabilities.
(3.37.c)
$$ \cptone{fair}{h,h,t}=\frac{\cpttwo{h,h,t}{fair}\prt{fair}}{\cpttwo{h,h,t}{fair}\prt{fair}+\cpttwo{h,h,t}{unfair}\prt{unfair}} $$
$$ =\frac{\frac12\frac12\frac12\frac12}{\frac12\frac12\frac12\frac12+0\wts1\wts1\wts\frac12}=1 $$
(3.38)
$$ \cp{T}{W}=\frac{\cp{W}{T}\pr{T}}{\cp{W}{T}\pr{T}+\cp{W}{H}\pr{H}}=\frac{\frac3{15}\frac12}{\frac3{15}\frac12+\frac5{12}\frac12} $$
$$ =\frac{\frac{12}{60}}{\frac{12}{60}+\frac{25}{60}}=\frac{12}{37} $$
(3.39)
This is a really important problem: unconditionally dependent and conditionally independent
Note that example 3a gives us
$$ \cp{A_1}{A}=\cp{A_2}{A}=0.4\dq\pr{A}=0.3\dq\pr{A_1}=0.26 $$
We can use Bayes to get
$$ \cp{A_2}{A_1^c}=\frac{\pr{A_1^cA_2}}{\pr{A_1^c}}=\frac{\cp{A_1^cA_2}{A}\pr{A}+\cp{A_1^cA_2}{A^c}\pr{A^c}}{\cp{A_1^c}{A}\pr{A}+\cp{A_1^c}{A^c}\pr{A^c}} \tag{3.39.1} $$
Short Answer:
This is a question of unconditionally dependent and conditionally independent:
When we don’t know whether the driver is accident prone, then $A_1$ and $A_2$ are dependent. For example: if we don’t know whether he is accident prone but we know that he had an accident in the first year, then we must assess higher** the probability of his getting into an accident in the second year. Hence unconditionally dependent. That is $\pr{A_1A_2}\neq\pr{A_1}\pr{A_2}$.
But if we know that he is or is not accident prone, this overrides the occurence of $A_1$ or $A_1^c$ and completely determines the probability that he will get into an accident in the second year. Hence conditionally independent. That is $\cp{A_1A_2}{A}=\cp{A_1}{A}\cp{A_2}{A}$.
So equation 3.39.1 becomes
$$ \frac{\cp{A_1^c}{A}\cp{A_2}{A}\pr{A}+\cp{A_1^c}{A^c}\cp{A_2}{A^c}\pr{A^c}}{1-\pr{A_1}} $$
$$ =\frac{0.6\wts0.4\wts0.3+0.8\wts0.2\wts0.7}{1-0.26} \tag{3.39.2} $$
In [953]: (.6*.4*.3+.8*.2*.7)/(1-.26)
Out[953]: 0.24864864864864863
** Define $A_u$ to be the updated accident proneness when there is an accident in the first year. Then
$$ \pr{A_u}=\cp{A}{A_1}=\frac{\cp{A_1}{A}\pr{A}}{\pr{A_1}}=\frac{0.4\wts0.3}{0.26}\approx0.4615384615384615>0.3=\pr{A} $$
Hence
$$ \pr{\overline{A}_{2}}=\cp{A_2}{A_u}\pr{A_u}+\cp{A_2}{A_u^c}\pr{A_u^c} $$
$$ =0.4\wts\pr{A_u}+0.2\wts(1-\pr{A_u})\approx0.2923076923076923 \tag{3.39.3} $$
Whereas
$$ \pr{A_2}=\cp{A_2}{A}\pr{A}+\cp{A_2}{A^c}\pr{A^c}=0.4\wts0.3+0.2\wts0.7=0.26 $$
So that
$$ \pr{\overline{A}_{2}}\approx0.2923>0.26=\pr{A_2} $$
That is, when we update the accident proneness with an accident in year $1$, the conditionally probability in year $2$ increases. This explains why we must assess higher the probability of an accident in year $2$ given an accident in year $1$. Notice that we can compute this more directly:
$$ \cp{A_2}{A_1}=\frac{\pr{A_1A_2}}{\pr{A_1}}=\frac{\cp{A_1A_2}{A}\pr{A}+\cp{A_1A_2}{A^c}\pr{A^c}}{0.26} $$
$$ =\frac{\cp{A_1}{A}\cp{A_2}{A}\pr{A}+\cp{A_1}{A^c}\cp{A_2}{A^c}\pr{A^c}}{0.26} $$
$$ =\frac{0.4\wts0.4\wts0.3+0.2\wts0.2\wts0.7}{0.26}\approx0.2923076923076923 $$
This answer matches exactly with 3.39.3. That is $\pr{\overline{A}_2}=\cp{A_2}{A_1}$.
We can also recompute $\cp{A_2}{A_1^c}$ the other way. Redefine $A_u$ to be the updated accident proneness when there is no accident in the first year. We should see that $\pr{A_u}<\pr{A}$:
$$ \pr{A_u}=\cp{A}{A_1^c}=\frac{\cp{A_1^c}{A}\pr{A}}{\pr{A_1^c}}=\frac{0.6\wts0.3}{0.74}\approx0.24324324324324323<0.3=\pr{A} $$
Hence
$$ \pr{\overline{A}_{2}}=\cp{A_2}{A_u}\pr{A_u}+\cp{A_2}{A_u^c}\pr{A_u^c} $$
$$ =0.4\wts\pr{A_u}+0.2\wts(1-\pr{A_u})\approx0.24864864864864866 $$
This answer matches with 3.39.2.
Longer Answer:
Let’s look at the term $\cp{A_1^cA_2}{A}$. Define $\prq{B}\equiv\cp{B}{A}$. Then $Q$ is a probability function and
$$ \cp{A_1^cA_2}{A}=\prq{A_1^cA_2}=\cpq{A_2}{A_1^c}\prq{A_1^c}=\cp{A_2}{A_1^cA}\cp{A_1^c}{A} $$
The equation $\cpq{A_2}{A_1^c}=\cp{A_2}{A_1^cA}$ follows from an unnumbered equation on p.94: when $Q$ is defined as $\prq{E}\equiv\cp{E}{F}$ then we can derive $\cpq{E_1}{E_2}=\cp{E_1}{E_2F}$.
We’ve been given no information in the problem or example 3a about accidents from year-to-year and their independence. So I guess we can assume that the probability of an accident in the second year given no accident in the first and given their accident prone is just the probability of an accident in a fixed year given their accident prone:
$$ \cp{A_2}{A_1^cA}=\cp{A_2}{A}=0.4 $$
The last equality was given in the example: An accident prone person has a probability of $0.4$ of getting into an accident within a fixed 1-year period. Hence
$$ \cp{A_1^cA_2}{A}=\cp{A_2}{A_1^cA}\cp{A_1^c}{A}=0.4\wts0.6 $$
We can handle $\cp{A_1^cA_2}{A^c}$ in a similiar way. Define $\prq{B}\equiv\cp{B}{A^c}$:
$$ \cp{A_1^cA_2}{A^c}=\prq{A_1^cA_2}=\cpq{A_2}{A_1^c}\prq{A_1^c}=\cp{A_2}{A_1^cA^c}\cp{A_1^c}{A^c} $$
And
$$ \cp{A_2}{A_1^cA^c}=\cp{A_2}{A^c}=0.2 $$
Hence
$$ \cp{A_1^cA_2}{A^c}=\cp{A_2}{A_1^cA^c}\cp{A_1^c}{A^c}=0.2\wts0.8 $$
So the nominator in equation 3.39.1 becomes
$$ \cp{A_1^cA_2}{A}\pr{A}+\cp{A_1^cA_2}{A^c}\pr{A^c}=0.4\wts0.6\wts0.3+0.2\wts0.8\wts0.7 $$
since we’re given $\pr{A}=0.3$. The denominator in equation 3.39.1 is
$$ \cp{A_1^c}{A}\pr{A}+\cp{A_1^c}{A^c}\pr{A^c}=0.6\wts0.3+0.8\wts0.7 $$
Hence equation 3.39.1 becomes
$$ \cp{A_2}{A_1^c}=\frac{0.4\wts0.6\wts0.3+0.2\wts0.8\wts0.7}{0.6\wts0.3+0.8\wts0.7} $$
In [720]: (.4*.6*.3+.2*.8*.7)/(.6*.3+.8*.7)
Out[720]: 0.24864864864864863
In [721]: 46/185 #solutions manual
Out[721]: 0.24864864864864866
3.40.a
There is $1=\binom30$ way to get no white balls:
$$ \pr{W_1^cW_2^cW_3^c}=\pr{W_1^c}\cp{W_2^c}{W_1^c}\cp{W_3^c}{W_1^cW_2^c}=\frac7{12}\frac8{13}\frac9{14}=\frac{2\wts9}{3\wts13\wts2}=\frac{3}{13} $$
3.40.b
There are $3=\binom31$ disjoint ways to get $1$ white ball, each with the same probability:
$$ \pr{W_1^cW_2^cW_3}=\pr{W_1^c}\cp{W_2^c}{W_1^c}\cp{W_3}{W_1^cW_2^c}=\frac7{12}\frac8{13}\frac5{14}=\frac{2\wts5}{3\wts13\wts2}=\frac{5}{39} $$
$$ \pr{W_1^cW_2W_3^c}=\pr{W_1^c}\cp{W_2}{W_1^c}\cp{W_3^c}{W_1^cW_2}=\frac7{12}\frac5{13}\frac8{14}=\frac{5}{39} $$
$$ \pr{W_1W_2^cW_3^c}=\frac{5}{39} $$
Hence $\prt{1 white ball}=3\wts\frac5{39}=\frac5{13}$.
3.40.c
There is $1=\binom33$ way to get $3$ white balls:
$$ \pr{W_1W_2W_3}=\pr{W_1}\cp{W_2}{W_1}\cp{W_3}{W_1W_2}=\frac5{12}\frac6{13}\frac7{14}=\frac{5}{2\wts13\wts2}=\frac{5}{52} $$
3.40.d
There are $3=\binom32$ disjoint ways to get $2$ white balls, each with the same probability:
$$ \pr{W_1^cW_2W_3}=\pr{W_1^c}\cp{W_2}{W_1^c}\cp{W_3}{W_1^cW_2}=\frac7{12}\frac5{13}\frac6{14}=\frac{5}{2\wts13\wts2}=\frac{5}{52} $$
$$ \pr{W_1W_2W_3^c}=\pr{W_1}\cp{W_2}{W_1}\cp{W_3^c}{W_1W_2}=\frac5{12}\frac6{13}\frac7{14}=\frac{5}{52} $$
$$ \pr{W_1W_2^cW_3}=\pr{W_1}\cp{W_2^c}{W_1}\cp{W_3}{W_1W_2^c}=\frac{5}{52} $$
Hence $\prt{2 white balls}=3\wts\frac5{52}=\frac{15}{52}$. And check
$$ \frac3{13}+\frac5{13}+\frac5{52}+\frac{15}{52}=\frac8{13}+\frac{20}{52}=\frac8{13}+\frac5{13} $$
3.41
????????????? come back and redo this. and save the python multinomial somewhere, either in previous problem or scratch.
Let $A$ denote the event that the second draw is an ace. Let $F$ denote the event that the interchanged ace is selected on the second draw. Conditioning on this:
$$ \pr{A}=\cp{A}{F}\pr{F}+\cp{A}{F^c}\pr{F^c}=1\wts\frac1{27}+\cp{A}{F^c}\frac{26}{27} $$
$F^c$ is the event that $1$ of the original $26$ cards in that half-deck is drawn second. So we want to condition on the number of aces in the original half-deck. Let $A_i$ denote the event that there are $i$ aces in the original half-deck:
$$ \cp{A}{F^c}=\sum_{i=0}^{3}\cp{A}{A_iF^c}\cp{A_i}{F^c} \tag{3.41.1} $$
$$ \cp{A_0}{F^c}=\pr{A_0}=\frac{\binom4{4,0}\binom{48}{24,24}}{\binom{52}{26,26}} $$
$$ \cp{A_1}{F^c}=\pr{A_1}=\frac{\binom4{3,1}\binom{48}{24,24}}{\binom{52}{26,26}} $$
$$ \cp{A_2}{F^c}=\pr{A_2}=\frac{\binom4{2,2}\binom{48}{24,24}}{\binom{52}{26,26}} $$
$$ \cp{A_3}{F^c}=\pr{A_3}=\frac{\binom4{1,3}\binom{48}{24,24}}{\binom{52}{26,26}} $$
# multinomial curtesy of https://stackoverflow.com/a/46374719/1884520
In [767]: from scipy.special import binom as sbinom
...:
...: def multinomial(params):
...: if len(params) == 1:
...: return 1
...: return sbinom(sum(params), params[-1]) * multinomial(params[:-1])
...:
In [768]: wmn=lambda lst:multinomial(lst)
In [751]: wmn([4,0])*wmn([24,24])/wmn([26,26])
Out[751]: 0.065026010404161699
In [752]: wmn([3,1])*wmn([24,24])/wmn([26,26])
Out[752]: 0.2601040416166468
In [753]: wmn([2,2])*wmn([24,24])/wmn([26,26])
Out[753]: 0.3901560624249702
In [754]: wmn([1,3])*wmn([24,24])/wmn([26,26])
Out[754]: 0.2601040416166468
In [755]: wmn([4,0])*wmn([24,24])/wmn([26,26])
Out[755]: 0.065026010404161699
Then we have
$$ \cp{A}{F^c}=\sum_{i=0}^{3}\cp{A}{A_iF^c}\cp{A_i}{F^c}=\sum_{i=0}^{3}\frac{i}{26}\frac{\binom4{4-i,i}\binom{48}{24,24}}{\binom{52}{26,26}} $$
In [765]: paifc=lambda i:wmn([4-i,i])*wmn([24,24])/wmn([26,26])
In [766]: sum([(i/26)*paifc(i) for i in range(0,4)])
Out[766]: 0.070028011204481835
$$ \pr{A}=1\wts\frac1{27}+\cp{A}{F^c}\frac{26}{27}\approx0.10447 $$
In [770]: (1/27)+sum([(i/26)*paifc(i) for i in range(0,4)])*(26/27)
Out[770]: 0.10447141819690843
$$ \cp{A}{F^c}=\sum_{i=0}^{3}\cp{A}{A_iF^c}\cp{A_i}{F^c}=\sum_{i=0}^{3}\cp{AA_i}{F^c}=0+ $$
(3.42)
Reworded question: If you get a cake from them, what is the probability that $A$ baked it, given that it’s a failed cake? Let $A$ denote the event that $A$ baked the cake. Let $F$ denote the event that it’s a failed cake.
$$ \cp{A}{F}=\frac{\cp{F}{A}\pr{A}}{\cp{F}{A}\pr{A}+\cp{F}{B}\pr{B}+\cp{F}{C}\pr{C}} $$
$$ =\frac{0.02\wts0.5}{0.02\wts0.5+0.03\wts0.3+0.05\wts0.2} $$
In [790]: (.02*.5)/(.02*.5+.03*.3+.05*.2)
Out[790]: 0.3448275862068965
In [791]: 10/29 #solutions manual
Out[791]: 0.3448275862068966
(3.43)
Let $A$ denote the event that it was the two-headed coin. Let $B$ denote the fair coin. Let $C$ denote the biased coin.
$$ \cp{A}{h}=\frac{\cp{h}{A}\pr{A}}{\cp{h}{A}\pr{A}+\cp{h}{B}\pr{B}+\cp{h}{C}\pr{C}} $$
$$ =\frac{1\wts\frac13}{1\wts\frac13+\frac12\frac13+\frac34\frac13}=\frac{1}{\frac44+\frac24+\frac34}=\frac1{\frac94}=\frac49 $$
(3.44)
Again, the wording of the problem leaves much to be desired. The author writes “Prisoner A asks the jailer to tell him privately which of his fellow prisoners will be set free”. It’s impossible to say what this means. Is the jailer to tell him both names if the other two are are to be set free? Or just one?
I think he means to say, “Prisoner A asks the jailer to tell him the name of one of his fellow prisoners who will be set free”.
Let $Z$ denote the event that the jailer tells prisoner A that prisoner B will be set free. Let $A$ denote the event that A dies, let $B$ denote the event that B dies, and let $C$ denote that C dies. Bayes Formula give us
$$ \cp{A}{Z}=\frac{\cp{Z}{A}\pr{A}}{\cp{Z}{A}\pr{A}+\cp{Z}{B}\pr{B}+\cp{Z}{C}\pr{C}} \tag{3.44.1} $$
If prisoner A is to die, then the jailer will tell prisoner A that either prisoner B or C will be set free. Presumably, it is equally likely that the jailer will tell him B or the jailer will tell him C. Hence $\cp{Z}{A}=\frac12$.
If prisoner B is to die, then the jailer will definitely tell prisoner A that prisoner C is to go free. But there’s no chance the jailer will tell him B is to go free. Hence $\cp{Z}{B}=0$.
If prisoner C is to die, the the jailer will definitely tell prisoner A that prisoner B is to go free. Hence $\cp{Z}{C}=1$.
Equation 3.44.1 becomes
$$ \cp{A}{Z}=\frac{\frac12\frac13}{\frac12\frac13+0\wts\frac13+1\wts\frac13}=\frac{\frac12}{\frac32}=\frac13 $$
We can do the same computation when $Z$ denotes the event that the jailer tell prisoner A that prisoner C is to be set free. Hence the jailer is wrong.
(3.45)
Let $C_i$ denote the event that the $i^{th}$ coin is selected.
$$ \cp{C_5}{h}=\frac{\cp{h}{C_5}\pr{C_5}}{\sum_{i=1}^{10}\cp{h}{C_i}\pr{C_i}}=\frac{\frac5{10}\frac1{10}}{\sum_{i=1}^{10}\frac{i}{10}\frac1{10}}=\frac{5}{\sum_{i=1}^{10}i}=\frac5{\frac{10\wts11}2}=\frac5{55}=\frac1{11} $$
(3.46)
Notice how similar this problem is to 3.39. They are actually identical with some letters and numbers replaced. Instead of $0.4=\cp{A_i}{A}$ and $0.2=\cp{A_i}{A^c}$, we are given $p_m=\cp{A_i}{M}$ and $p_f=\cp{A_i}{F}$. Initially, I really struggled with why $A_1$ and $A_2$ are unconditionally dependent and conditionally independent. But my answer to 3.39 explains this in detail.
Let $M$ denote the event that the policyholder is male. Then
$$ \cp{A_2}{A_1}=\frac{\pr{A_2A_1}}{\pr{A_1}}=\frac{\cp{A_1A_2}{M}\pr{M}+\cp{A_1A_2}{F}\pr{F}}{\cp{A_1}{M}\pr{M}+\cp{A_1}{F}\pr{F}} $$
$$ =\frac{\cp{A_1}{M}\cp{A_2}{M}\pr{M}+\cp{A_1}{F}\cp{A_2}{F}\pr{F}}{p_m\alpha+p_f(1-\alpha)} $$
$$ =\frac{p_m^2\alpha+p_f^2(1-\alpha)}{p_m\alpha+p_f(1-\alpha)} $$
So we want to show that
$$ \frac{p_m^2\alpha+p_f^2(1-\alpha)}{p_m\alpha+p_f(1-\alpha)}=\cp{A_2}{A_1}>\pr{A_1}=p_m\alpha+p_f(1-\alpha) $$
$\iff$
$$ p_m^2\alpha+p_f^2(1-\alpha)>\bop p_m\alpha+p_f(1-\alpha)\bcp^2 $$
$\iff$
$$ p_m^2\alpha+p_f^2(1-\alpha)>p_m^2\alpha^2+p_f^2(1-\alpha)^2+2p_mp_f\alpha(1-\alpha) $$
$\iff$
$$ p_m^2\alpha-p_m^2\alpha^2+p_f^2(1-\alpha)-p_f^2(1-\alpha)^2>2p_mp_f\alpha(1-\alpha) $$
$\iff$
$$ p_m^2\bop\alpha-\alpha^2\bcp+p_f^2\bop(1-\alpha)-(1-\alpha)^2\bcp>2p_mp_f\alpha(1-\alpha) \tag{3.46.1} $$
But
$$ 1-\alpha-(1-\alpha)^2=1-\alpha-(1-2\alpha+\alpha^2)=1-\alpha-1+2\alpha-\alpha^2 $$
$$ =\alpha-\alpha^2=\alpha(1-\alpha) $$
So we factor $\alpha(1-\alpha)$ out of 3.46.1 to get
$$ p_m^2+p_f^2>2p_mp_f $$
$\iff$
$$ (p_m-p_f)^2=p_m^2-2p_mp_f+p_f^2>0 $$
$\iff$
$$ p_m\neq p_f $$
which was given in the problem statement.
(3.47)
Let $W$ denote the event that all balls selected are white. Let $D_i$ denote the event that the die landed on $i$.
We two alternatives for computing this:
$$ \pr{W}=\sum_{i=1}^{6}\cp{W}{D_i}\pr{D_i}=\sum_{i=1}^{6}\frac{\binom5i}{\binom{15}i}\frac16=\frac16\sum_{i=1}^{6}\frac{\binom5i}{\binom{15}i} $$
In [992]: (1/6)*sum([winom(5,i)/winom(15,i) for i in range(1,7)])
Out[992]: 0.0757575757575757
or
$$ \pr{W}=\sum_{i=1}^{6}\cp{W}{D_i}\pr{D_i}=\sum_{i=1}^{6}\Bop\prod_{j=0}^{i-1}\frac{5-j}{15-j}\Bcp\frac16=\frac16\sum_{i=1}^{6}\Bop\prod_{j=0}^{i-1}\frac{5-j}{15-j}\Bcp $$
In [812]: (1/6)*sum([prod([(5-j)/(15-j) for j in range(0, i)]) for i in range(1,7)])
Out[812]: 0.075757575757575746
$$ \cp{D_3}{W}=\frac{\cp{W}{D_3}\pr{D_3}}{\pr{W}}=\frac{\frac5{15}\frac4{14}\frac3{13}\frac16}{\pr{W}} $$
In [814]: ((5/15)*(4/14)*(3/13)*(1/6))/wp
Out[814]: 0.04835164835164836
(3.48)
Let $S_i$ denote the event that a silver coin is selected on the $i^{th}$ pick. Let $A$ denote the event that cabinet A was selected and let $B$ denote the event that cabinet B was selected.
$$ \cp{S_2}{S_1}=\frac{\pr{S_1S_2}}{\pr{S_1}}=\frac{\cp{S_1S_2}{A}\pr{A}+\cp{S_1S_2}{B}\pr{B}}{\cp{S_1}{A}\pr{A}+\cp{S_1}{B}\pr{B}}=\frac{1\wts\frac12+0\wts\frac12}{1\wts\frac12+\frac12\frac12}=\frac{\frac24}{\frac34}=\frac23 $$
Alternative computation: Note that there will be a silver coin in the other drawer IFF cabinet $A$ was selected.
$$ \cp{S_2}{S_1}=\cp{A}{S_1}=\frac{\cp{S_1}{A}\pr{A}}{\cp{S_1}{A}\pr{A}+\cp{S_1}{B}\pr{B}}=\frac{1\wts\frac12}{1\wts\frac12+\frac12\frac12}=\frac{\frac12}{\frac34}=\frac23 $$
(3.49.a)
Let $C$ denote the event that the male has cancer. Let $E$ denote the event of an elevated PSA level. We’re given that
- $\pr{C}=0.7$
- $\cp{E}{C^c}=0.135$
- $\cp{E}{C}=0.268$
$$ \cp{C}{E}=\frac{\cp{E}{C}\pr{C}}{\cp{E}{C}\pr{C}+\cp{E}{C^c}\pr{C^c}}=\frac{0.268\wts0.7}{0.268\wts0.7+0.135\wts0.3} $$
In [825]: pce=lambda ip,gn,gc:(gc*ip)/(gc*ip+gn*(1-ip))
In [826]: pce(.7,.135,.268)
Out[826]: 0.8224462954844366
In [827]: pce(.3,.135,.268)
Out[827]: 0.45969125214408235
(3.49.b)
Since conditional probabilities are probabilities, note that
- $\cp{E^c}{C^c}=1-\cp{E}{C^c}=1-0.135=0.865$
- $\cp{E^c}{C}=1-\cp{E}{C}=1-0.268=0.732$
$$ \cp{C}{E^c}=\frac{\cp{E^c}{C}\pr{C}}{\cp{E^c}{C}\pr{C}+\cp{E^c}{C^c}\pr{C^c}}=\frac{0.732\wts0.7}{0.732\wts0.7+0.865\wts0.3} $$
In [828]: pce(.7,.865,.732)
Out[828]: 0.6638165565487757
In [829]: pce(.3,.865,.732)
Out[829]: 0.26614955762937825
(3.50)
Select a random person. Let $A$ denote the event that the person has an accident in a fixed year. Let $G, M, B$ denote the events that the person’s risk level is good, average, bad.
$$ \pr{A}=\cp{A}{G}\pr{G}+\cp{A}{M}\pr{M}+\cp{A}{B}\pr{B}=0.05\wts0.2+0.15\wts0.5+0.3\wts0.3 $$
In [830]: .05*.2+.15*.5+.3*.3
Out[830]: 0.175
Note that $G$ and $M$ are disjoint and that $\cp{\cdot}{A^c}$ is a probability function. So
$$ \cp{G\cup M}{A^c}=\cp{G}{A^c}+\cp{M}{A^c}=\frac{\cp{A^c}{G}\pr{G}+\cp{A^c}{M}\pr{M}}{\pr{A^c}} $$
$$ =\frac{(1-0.05)0.2+(1-0.15)0.5}{1-0.175} $$
In [833]: ((1-.05)*.2+(1-.15)*.5)/(1-.175)
Out[833]: 0.7454545454545455
Or
$$ \cp{G\cup M}{A^c}=1-\cp{B}{A^c}=1-\frac{\cp{A^c}{B}\pr{B}}{\pr{A^c}}=1-\frac{(1-0.3)0.3}{1-0.175} $$
In [834]: 1-((1-.3)*.3)/(1-.175)
Out[834]: 0.7454545454545454
(3.51.a)
She is $65\%$ certain:
In [837]: .7*.8+.2*.4+.1*.1
Out[837]: 0.6499999999999999
More formally, let $J$ denote the event that she gets the job, and let $S,G,W$ denote the events that she receives strong, good, and weak recommendations.
$$ \pr{J}=\cp{J}{S}\pr{S}+\cp{J}{G}\pr{G}+\cp{J}{W}\pr{W}=0.8\wts0.7+0.4\wts0.2+0.1\wts0.1 $$
(3.51.b)
$$ \cp{S}{J}=\frac{\cp{J}{S}\pr{S}}{\pr{J}}=\frac{0.8\wts0.7}{0.65} $$
$$ \cp{G}{J}=\frac{\cp{J}{G}\pr{G}}{\pr{J}}=\frac{0.4\wts0.2}{0.65} $$
$$ \cp{W}{J}=\frac{\cp{J}{W}\pr{W}}{\pr{J}}=\frac{0.1\wts0.1}{0.65} $$
In [886]: jo=lambda jc,p,c=False: (jc*p)/(c+pw(-1,c)*(.7*.8+.2*.4+.1*.1))
In [887]: jo(.8,.7),jo(.4,.2),jo(.1,.1)
Out[887]: (0.86153846153846159, 0.12307692307692311, 0.015384615384615389)
(3.51.c)
$$ \cp{S}{J^c}=\frac{\cp{J^c}{S}\pr{S}}{\pr{J^c}}=\frac{0.2\wts0.7}{0.35} $$
$$ \cp{G}{J^c}=\frac{\cp{J^c}{G}\pr{G}}{\pr{J^c}}=\frac{0.6\wts0.2}{0.35} $$
$$ \cp{W}{J^c}=\frac{\cp{J^c}{W}\pr{W}}{\pr{J^c}}=\frac{0.9\wts0.1}{0.35} $$
In [888]: jo(.2,.7,c=True),jo(.6,.2,c=True),jo(.9,.1,c=True)
Out[888]: (0.39999999999999986, 0.34285714285714275, 0.25714285714285712)
(3.52.a)
Let $M,T,W,Z,F$ denote the events that she receives mail on Monday, Tuesday, Wednesday, Thursday, or Friday, respectively. And let $A$ denote the event that she’s accepted and let $R$ denote the event that she’s rejected.
$$ \pr{M}=\cp{M}{A}\pr{A}+\cp{M}{R}\pr{R}=0.15\wts0.6+0.05\wts0.4=0.09+0.02=0.11 $$
$$ \pr{T}=\cp{T}{A}\pr{A}+\cp{T}{R}\pr{R}=0.2\wts0.6+0.1\wts0.4=0.12+0.04=0.16 $$
(3.52.b)
$$ \cp{T}{M^c}=\frac{\pr{TM^c}}{\pr{M^c}}=\frac{\pr{T}}{1-0.11}=\frac{0.16}{0.89}\approx0.179775 $$
(3.52.c)
Let $S$ denote the union of every day other than listed. Note that
$$ M^c=T\cup W\cup Z\cup F\cup S\dq T^c=M\cup W\cup Z\cup F\cup S\dq W^c=M\cup T\cup Z\cup F\cup S $$
Hence
$$ M^cT^cW^c=Z\cup F\cup S $$
And
$$ \cp{A}{M^cT^cW^c}=\cp{A}{Z\cup F\cup S}=\frac{\cp{Z\cup F\cup S}{A}\pr{A}}{\pr{Z\cup F\cup S}} $$
$$ =\frac{\bop\cp{Z}{A}+\cp{F}{A}+\cp{S}{A}\bcp\pr{A}}{\pr{Z}+\pr{F}+\pr{S}} $$
$$ =\frac{\bop0.15+0.1+0.15\bcp0.6}{0.15\wts0.6+0.15\wts0.4+0.1\wts0.6+0.2\wts0.4+0.15\wts0.6+0.4\wts0.4} $$
In [1000]: ac=lambda o=.2,t=.1: o*.6+t*.4
In [1001]: ((.15+.1+.15)*.6)/(ac(.15,.15)+ac(.1,.2)+ac(.15,.4))
Out[1001]: 0.4444444444444444
Alternative computation:
$$ \cp{A}{M^cT^cW^c}=\frac{\cp{M^cT^cW^c}{A}\pr{A}}{\cp{M^cT^cW^c}{A}\pr{A}+\cp{M^cT^cW^c}{R}\pr{R}} \tag{3.52.c.1} $$
Aside: $A^cB^c=(A\cup B)^c$
Proof: Suppose $x\in A^cB^c$. Then $x\in A^c$ and $x\in B^c$. Then $x\not\in A$ and $x\not\in B$. So $x\not\in A\cup B$. So $x\in(AUB)^c$. Conversely, suppose $x\in(A\cup B)^c$. Then $x\not\in A\cup B$. Then $x\not\in A$ and $x\not\in B$. So $x\in A^c$ and $x\in B^c$. So $x\in A^cB^c$. $\wes$
Now let’s compute the first term in the denominator in equation 3.52.c.1:
$$ \cp{M^cT^cW^c}{A}=\cp{(M\cup T\cup W)^c}{A} \tag{Aside} $$
$$ =1-\cp{M\cup T\cup W}{A} \tag{Complement} $$
$$ =1-\bop\cp{M}{A}+\cp{T}{A}+\cp{W}{A}\bcp \tag{Mutually Exclusive} $$
$$ =1-(0.15+0.20+0.25)=0.4 $$
Similarly
$$ \cp{M^cT^cW^c}{R}=1-\cp{M\cup T\cup W}{R}=1-(0.05+0.1+0.1)=0.75 $$
Equation 3.52.c.1 becomes
$$ \cp{A}{M^cT^cW^c}=\frac{0.4\wts0.6}{0.4\wts0.6+0.75\wts0.4}=\frac{0.24}{0.54}=\frac49=0.\overline{4} $$
(3.52.d)
$$ \cp{A}{Z}=\frac{\cp{Z}{A}\pr{A}}{\cp{Z}{A}\pr{A}+\cp{Z}{R}\pr{R}}=\frac{0.15\wts0.6}{0.15\wts0.6+0.15\wts0.4}=0.6 $$
(3.52.e)
$$ \cp{A}{S}=\frac{\cp{S}{A}\pr{A}}{\cp{S}{A}\pr{A}+\cp{S}{R}\pr{R}}=\frac{0.15\wts0.6}{0.15\wts0.6+0.4\wts0.4}=\frac{0.09}{0.09+0.16}=0.36 $$
Alternative computation: In part (c), we showed that $M^cT^cW^cZ^cF^c=(M\cup T\cup W\cup Z\cup F)^c$ and
$$ \cp{M^cT^cW^cZ^cF^c}{A}=\cp{(M\cup T\cup W\cup Z\cup F)^c}{A} $$
$$ =1-\cp{M\cup T\cup W\cup Z\cup F}{A} $$
$$ =1-\bop\cp{M}{A}+\cp{T}{A}+\cp{W}{A}+\cp{Z}{A}+\cp{F}{A}\bcp $$
$$ =1-(0.15+0.20+0.25+0.15+0.10)=0.15 $$
Similarly
$$ \cp{M^cT^cW^cZ^cF^c}{R}=1-(0.05+0.10+0.10+0.15+0.20)=0.4 $$
Hence
$$ \cpttwo{A}{no mail this week}=\cp{A}{M^cT^cW^cZ^cF^c} $$
$$ =\frac{\cp{M^cT^cW^cZ^cF^c}{A}\pr{A}}{\cp{M^cT^cW^cZ^cF^c}{A}\pr{A}+\cp{M^cT^cW^cZ^cF^c}{R}\pr{R}} $$
$$ =\frac{0.15\wts0.6}{0.15\wts0.6+0.4\wts0.4}=\frac{0.09}{0.09+0.16}=\frac9{25}=0.36 $$
(3.53)
Let $C_i$ denote the event that component $i$ is functioning. Let $F$ denote the event that that system is functioning.
If we’re not careful, we can easily compute the wrong answer:
$$ \cp{C_1}{F}=\frac{\cp{F}{C_1}\pr{C_1}}{\sum_{i=1}^{n}\cp{F}{C_i}\pr{C_i}}=\frac{1\wts\frac12}{\sum_{i=1}^{n}1\wts\frac12}=\frac{\frac12}{\frac{n}2}=\frac1n $$
This answer is wrong because the $\set{C_i}$ are not mutually exclusive.
Note that $F=\cup_iC_i$ so that $C_i\subset F$. Also note that $F$ is a binomial random variable:
$$ \pr{F}=\sum_{i=1}^n\binom{n}i\Bop\frac12\Bcp^{i}\Bop\frac12\Bcp^{n-i} $$
But it’s easier compute the complement:
$$ \pr{F}=1-\pr{F^c}=1-\prt{zero components work}=1-\binom{n}0\Bop\frac12\Bcp^{0}\Bop\frac12\Bcp^{n-0}=1-\Bop\frac12\Bcp^{n} $$
Check:
In [1006]: brv=lambda n=10,i=0,p=1/2:winom(n,i)*pw(p,i)*pw(1-p,n-i)
In [1007]: brv()
Out[1007]: 0.000976562500000000
In [1008]: 1-sum([brv(10,i) for i in range(1,11)])
Out[1008]: 0.000976562500000000
Hence
$$ \cp{C_1}{F}=\frac{\pr{C_1F}}{\pr{F}}=\frac{\pr{C_1}}{1-\pr{F^c}}=\frac{\frac12}{1-\bop\frac12\bcp^n} $$
(3.54)
I would prefer to answer these questions:
In parts (a) and (b), identify whether the events are disjoint, independent, or neither (events cannot be both disjoint and independent).
- You and a randomly selected student from your class both earn A’s in this course.
- You and your class study partner both earn A’s in this course.
- If two events can occur at the same time, must they be dependent?
Solutions:
a. If the class is not graded on a curve, they are independent and not disjoint. If graded on a curve, they’re dependent. That is, only so many students will get an A. So my getting an A is dependent on him getting an A. If graded on a curve, they’re not disjoint unless the instructor will only give one A. In that case they’re very much disoint.
This is a good time to introduce weak dependence. Suppose there is a curve and that the class size is $n\equiv20$. Also suppose that $p\equiv15\%$ of the students will get an A in this curve. Let $M$ denote the event that I get an A. Then the unconditional probability that I will get an A is $\pr{M}=15\%$. This can also be computed as a hypergeometric random variable:
$$ \pr{M}=\frac{\binom11\binom{19}2}{\binom{20}{3}}=\frac{\binom{n-1}{np-1}}{\binom{n}{np}} \tag{3.54.1} $$
In [1141]: uncond=lambda n=100,p=0.15:(winom(n-1,n*p-1))/(winom(n,n*p))
In [1142]: uncond(20)
Out[1142]: 0.150000000000000
Let $B$ denote the event that the other randomly chosen student gets an A. Then
$$ \cp{M}{B}=\frac{\binom{18}1}{\binom{19}{2}}=\frac{\binom{n-2}{np-2}}{\binom{n-1}{np-1}} \tag{3.54.2} $$
In [1143]: cond=lambda n=100,p=0.15:(winom(n-2,n*p-2)/winom(n-1,n*p-1))
In [1144]: cond(20)
Out[1144]: 0.105263157894737
Notice that the unconditional probability is quite a bit larger than the conditional probability. So $M$ and $B$ are dependent. We say that two events are weakly dependent if the conditional probability converges to the unconditional probability as the sample size $n$ gets large.
In [1145]: weak_dep=lambda n=100,p=0.15:(uncond(n,p),cond(n,p))
In [1146]: weak_dep(20)
Out[1146]: (0.150000000000000, 0.105263157894737)
In [1147]: weak_dep(40)
Out[1147]: (0.150000000000000, 0.128205128205128)
In [1148]: weak_dep(80)
Out[1148]: (0.150000000000000, 0.139240506329114)
In [1149]: weak_dep(160)
Out[1149]: (0.150000000000000, 0.144654088050314)
In [1150]: weak_dep(320)
Out[1150]: (0.150000000000000, 0.147335423197492)
In [1151]: weak_dep(640)
Out[1151]: (0.150000000000000, 0.148669796557120)
Hence $M$ and $B$ are weakly dependent.
b. They are probably dependent: if we study together, our study habits would be related, which suggests our course performances are also related. They’re not disjoint unless the instructor will only give one A. In that case, they’re very much disoint.
c. No. See the answer to part (a) when the course is not graded on a curve. More generally: if two things are unrelated (independent), then one occurring does not preclude the other from occurring.
(3.55)
Independence means that $\pr{AB}=\pr{A}\pr{B}$. Let $n$ denote the number of sophomore girls present in class.
$$ \pr{\text{boy}\cap\text{fresh}}=\frac4{16+n}\dq\prt{boy}=\frac{10}{16+n}\dq\prt{fresh}=\frac{10}{16+n} $$
So we want to select $n$ such that $\pr{\text{boy}\cap\text{fresh}}=\prt{boy}\prt{fresh}$:
$$ \frac4{16+n}=\frac{10}{16+n}\frac{10}{16+n}=\frac{100}{(16+n)^2} $$
$\iff$
$$ 4=\frac{100}{16+n}\iff4n+64=100\iff4n=36\iff n=9 $$
Let’s verify this:
$$ \pr{\text{boy}\cap\text{fresh}}=\frac4{25}\dq\dq\prt{boy}\prt{fresh}=\frac{10}{25}\frac{10}{25}=\frac{4\wts25}{25\wts25} $$
Or
$$ \pr{\text{girl}\cap\text{soph}}=\frac9{25}\dq\dq\prt{girl}\prt{soph}=\frac{15}{25}\frac{15}{25}=\frac{5\wts3\wts5\wts3}{25\wts25} $$
(3.56)
Let $T_i$ denote the event that the just-obtained coupon is of type $i$. Let $N$ denote the event that the just-obtained coupon is a new type for our collection.
$$ \pr{N}=\sum_{i=1}^{m}\cp{N}{T_i}\pr{T_i} $$
Let’s start with type $1$. This just-obtained coupon is of type $1$ with probability $\pr{T_1}=p_1$.
Now let’s look at $\cp{N}{T_1}$. Type $1$ will be a new type for our $n-1$ collection IFF every one of these $n-1$ coupons is NOT of type $1$. Let $Z_j$ denote the event that coupon $j$ in our $n-1$ collection is not of type $1$. Then
$$ \cset{N}{T_1}=\cap_{j=1}^{n-1}Z_j $$
If we assume that coupon types are collected independently, then
$$ \cp{N}{T_1}=\pr{\cap_{j=1}^{n-1}Z_j}=\prod_{j=1}^{n-1}\pr{Z_j} $$
Each of these $n-1$ coupons is of type $1$ with probability $p_1$. So each of these $n-1$ coupons is NOT of type $1$ with probability $1-p_1$. That is, $\pr{Z_j}=1-p_1$ for $j=1,\dots,n-1$. Hence
$$ \cp{N}{T_1}=\pr{\cap_{j=1}^{n-1}Z_j}=\prod_{j=1}^{n-1}\pr{Z_j}=(1-p_1)^{n-1} $$
Similar conclusions can be reached for each event $T_i$, $i=1,…,m$. Hence
$$ \pr{N}=\sum_{i=1}^{m}\cp{N}{T_i}\pr{T_i}=\sum_{i=1}^{m}(1-p_i)^{n-1}p_i $$
(3.57.a)
There are two ways we can return to the original price: $U_1D_2$ and $D_1U_2$. The other two arrangements $U_1U_2$ and $D_1D_2$ don’t return to the original price.
Since changes on different days are assumed to be independent, we have
$$ \pr{U_1D_2}=\pr{U_1}\pr{D_2}=p(1-p) $$
Same calculation for $D_1U_2$.
Also notice that $U_1D_2$ and $D_1U_2$ are mutually exclusive.
$$ \prt{orig price after 2 days}=\pr{U_1D_2\cup D_1U_2}=\pr{U_1D_2}+\pr{D_1U_2}=2p(1-p) $$
(3.57.b)
The sample space consists of $8=2^3$ arrangements. And there are $\binom32$ ways to arrange two u’s and a d. The probability is the same for each combination.
$$ \pr{U_1U_2D_3}=\pr{U_1}\pr{U_2}\pr{D_3}=p^2(1-p) $$
Each such combination is disjoint from the others. Hence
$$ \prt{up 1 after 3 days}=\binom32p^2(1-p)=3p^2(1-p) $$
(3.57.c)
Let $Z_{1,3}$ denote the event that the stock’s price is up $1$ unit after $3$ days. In part (b), we showed that $\pr{Z_{1,3}}=3p^2(1-p)$.
Let $F_{i,j}$ denote the event that the stock’s price is flat from the end of day $i$ to the end of day $j$. In part (a), we computed $\pr{F_{1,3}}=2p(1-p)$.
Notice that
$$ U_1Z_{1,3}=\set{(u,u,d),(u,d,u)}=U_1F_{1,3} $$
Also notice that $U_1$ and $F_{1,3}$ are independent since changes on different days are assumed to be independent. Hence
$$ \cp{U_1}{Z_{1,3}}=\frac{\pr{U_1Z_{1,3}}}{\pr{Z_{1,3}}}=\frac{\pr{U_1}\pr{F_{1,3}}}{3p^2(1-p)}=\frac{p2p(1-p)}{3p^2(1-p)}=\frac23 $$
Or
$$ \cp{U_1}{Z_{1,3}}=\frac{\wN{U_1Z_{1,3}}}{\wN{Z_{1,3}}}=\frac{\wN{U_1}\wN{F_{1,3}}}3=\frac{1\wts2}3=\frac23 $$
Since $Z_{1,3}=\set{(u,u,d),(u,d,u),(d,u,u)}$.
(3.58.a)
$$ \underbrace{h,h}_{N},\underbrace{t,t}_{N},\underbrace{t,t}_{N},\underbrace{h,h}_{N},\underbrace{h,t}_{T} \tag{3.58.1} $$
Since the outcomes $(h,h)$ and $(t,t)$ amount to no-flips, we can disregard them. We are only concerned with $H=(t,h)$ and $T=(h,t)$. We want to show $(t,h)$ will occur with the same long-run frequency as $(h,t)$. That is, we want to show that $\pr{(t,h)}=\pr{(h,t)}$. Presumably the biased coin is independent:
$$ \pr{(t,h)}=\pr{t}\pr{h}=\pr{(h,t)} $$
Alternatively, we can compare the frequencies of $H=(t,h)$ and $H\cup T=(t,h)\cup(h,t)$. That is, when $H\cup T$ occurs, $H$ should then occur half the time. We have
$$ \pr{H}=\cp{(t,h)}{(t,h)\cup(h,t)}=\frac{\prb{(t,h)\cap\bop(t,h)\cup(h,t)\bcp}}{\prb{(t,h)\cup(h,t)}} $$
$$ =\frac{\pr{(t,h)}}{\pr{(t,h)}+\pr{(h,t)}}=\frac{p(1-p)}{p(1-p)+(1-p)p}=\frac{p(1-p)}{2p(1-p)}=\frac12 $$
(3.58.b)
Analyze this procedure: continue to flip the coin until the last two flips are different and then let the result be the outcome of the final flip. Analyzing this procedure on the same sequence from 3.58.1, we see that
$$ \underbrace{h,h,t}_{T},\underbrace{t,t,t,h}_{H},\underbrace{h,h,t}_{T} \tag{3.58.1} $$
Whereas the procedure in part (a) consists of outcomes $(h,h),(t,t)$ that are irrelevent to the result, every outcome matters in this procedure.
Heads occurs by flipping an arbitrary number of tails and finally flipping heads. That is, heads occurs by first flipping tails. Hence $\pr{H}=\pr{t}=1-p$. Similarly $\pr{T}=\pr{h}=p$. This will not be a fair coin unless $p=\frac12$.
(3.59.a)
$p^4$
(3.59.b)
$(1-p)p^3$
(3.59.c)
The trick is that $H,H,H,H$ must occur at the very beginning of the sequence if it’s going to precede $T,H,H,H$. For instance:
$$ H,H,T,H,T,\underbrace{T,H,H,H}_{THHH},H $$
We see that the event $T,H,H,H$ has occurred. However, also notice that the event $H,H,H,H$ has occurred. In order for the event $H,H,H,H$ to occur it must either occur at the very beginning of the sequence or it must be preceded by the event $T,H,H,H$. Hence
$$ \pr{THHH\text{ before }HHHH}=1-\pr{HHHH\text{ before }THHH} $$
$$ =1-\pr{HHHH\text{ occurs at beginning of seq}}=1-p^4 $$
(3.60.a)
Let’s first note that each parent must possess exactly $1$ blue-eyed gene and $1$ brown-eyed gened. Suppose not. Suppose the dad has $2$ blue-eyed genes. Then he has blue eyes. Contradiction. Now suppose the dad has $2$ brown-eyed genes. Then Smith’s sister must get at least brown-eyed gene. So she must have brown eyes. Contradiction. So it must be that the dad has exactly $1$ blue-eyed gene and $1$ brown-eyed gened. Similar reasoning shows that the mom must have exactly $1$ blue-eyed gene and $1$ brown-eyed gened as well.
For Smith to possess a blue-eyed gene, he must possess exactly $1$. Obviously he can’t have $2$ since then he would have blue eyes.
But let’s ignore this fact for a moment. Let $X$ denote the number of blue-eyed genes that Smith possesses. Let $\set{j,k}$ denote the event (i.e. sequence) that Smith acquired gene $j$ from his dad and gene $k$ from his mom. Presumably this is an independent sequence. Hence $\pr{j,k}=\pr{j}\pr{k}$. And let $B$ denote a brown-eyed gene and $b$ denote a blue-eyed gene.
There are two mutually exclusive ways that Smith can get exactly $1$ blue-eyed gene: $\set{b,B}$ or $\set{B,b}$. In the first sequence, the probability is $\pr{b,B}=\pr{b}\pr{B}=\frac12\frac12=\frac14$. Similarly $\pr{B,b}=\frac14$. Hence $\pr{X=1}=2\wts\frac14=\frac12$. Of course $X$ is a binomial random variable: $\pr{X=1}=\binom21\frac12\frac12=\frac12$
Similary we see that $\pr{X=0}=\binom20\bop\frac12\bcp^2=\frac14$ and $\pr{X=2}=\binom22\bop\frac12\bcp^2=\frac14$.
But we also know that Smith has brown eyes and hence doesn’t have $2$ blue-eyed genes. That is, $X<2$:
$$ \cpttwo{X=1}{Smith has brown eyes}=\cp{X=1}{X<2} $$
$$ =\frac{\pr{X=1\cap X<2}}{\pr{X<2}}=\frac{\pr{X=1}}{1-\pr{X=2}}=\frac{\frac12}{1-\frac14}=\frac23 $$
(3.60.b)
Let $F_i$ denote the event that their $i^{th}$ child will have blue eyes. Let $S$ denote the event that Smith has exactly $1$ blue-eyed gene.
$$ \pr{F_1}=\cp{F_1}{S}\pr{S}+\cp{F_1}{S^c}\pr{S^c}=\frac12\frac23+0\wts\frac12=\frac13 $$
(3.60.c)
First we need to update the conditional probability that Smith has a blue-eyed gene from the information that their first child has brown eyes. Before we do that, let’s compute:
$$ \cp{F_1^c}{S}=\frac{\wN{(B,b_1),(B,b_2)}}{\wN{(B,b_1),(B,b_2),(b,b_1),(b,b_2)}}=\frac24=\frac12 $$
since each outcome is mutually exclusive and equally likely to occur. Since this is an independent sequence, we also could have computed this as: $\cp{F_1^c}{S}=2\wts\frac12\frac12=2\wts\frac14=\frac12$.
Now let $S_u$ denote the updated event that Smith has a blue-eyed gene and compute its probability:
$$ \pr{S_u}=\cp{S}{F_1^c}=\frac{\cp{F_1^c}{S}\pr{S}}{\pr{F_1^c}}=\frac{\frac12\frac23}{1-\pr{F_1}}=\frac{\frac12\frac23}{\frac23}=\frac12 $$
Note that the occurence of $F_1^c$ implies a different probability that Smith has a blue-eyed gene. Hence one way to compute $\cp{F_2^c}{F_1^c}$ is to condition $F_2^c$ on $S_u$:
$$ \cp{F_2^c}{F_1^c}=\cp{F_2^c}{S_u}\pr{S_u}+\cp{F_2^c}{S_u^c}\pr{S_u^c} $$
$$ =\Bocb2\wts\frac12\frac12\Bccb\frac12+1\wts\frac12=\frac14+\frac12=\frac34 $$
There is an alternative way to compute this conditional probability. I think it is more direct: $\cp{F_2^c}{F_1^c}=\frac{\pr{F_2^cF_1^c}}{\pr{F_1^c}}$. But we must be careful. We cannot simply compute $\pr{F_2^cF_2^c}=\pr{F_2^c}\pr{F_1^c}$. This implies $F_1$ and $F_2$ are independent. But they are not. If we don’t take into account the genes of Smith, then the occurence of $F_1$ or $F_1^c$ will imply a different probability for the occurence of $F_2$.
Said another way: suppose we don’t know anything about Smith and his genes. And suppose that the first child is born with blue eyes. Then the conditional probability that the second child will be born with blue eyes rises considerably. Hence they are not independent events.
But these events are conditionally independent if we know how many blue-eyed genes Smith has. In this case, we are back to square $1$. That is, we know exactly which genes the parents have and so all possible gene sequences are independent from the others. Said another way, if the genes of both parents are known, then the eye color of the first child is irrelevent when computing the probability of the eye color of the second child. Hence:
$$ \cp{F_2^c}{F_1^c}=\frac{\pr{F_2^cF_1^c}}{\pr{F_1^c}}=\frac{\cp{F_2^cF_1^c}{S}\pr{S}+\cp{F_2^cF_1^c}{S^c}\pr{S^c}}{1-\pr{F_1}} $$
$$ =\frac{\cp{F_2^c}{S}\cp{F_1^c}{S}\pr{S}+\cp{F_2^c}{S^c}\cp{F_1^c}{S^c}\pr{S^c}}{1-\frac13} \tag{Independence} $$
$$ =\frac{\frac12\frac12\frac23+1\wts1\wts\frac13}{\frac23}=\frac32\Bop\frac{2}{12}+\frac13\Bcp=\frac32\Bop\frac16+\frac26\Bcp=\frac32\frac12=\frac34 $$
(3.61.a)
Let $C$ denote the event that the nonalbino kid is a carrier. Let $X$ denote the number of albino genes ‘a’ that the nonalbino kid has. Then $\pr{X=1}=2\wts\frac12\frac12=\frac12$ and
$$ \pr{C}=\cp{X=1}{X<2}=\frac{\pr{X=1\cap X<2}}{\pr{X<2}}=\frac{\pr{X=1}}{1-\pr{X=2}}=\frac{\frac12}{1-\frac14}=\frac23 $$
Let $A_i$ denote the event that the $i^{th}$ offspring is albino. Let $Y$ denote the number of albino genes that their first off-spring will have.
$$ \prt{first off-spring albino}=\pr{A_1}=\pr{Y=2} $$
$$ =\cp{Y=2}{C}\pr{C}+\cp{Y=2}{C^c}\pr{C^c}=\frac14\frac23+0=\frac16 $$
(3.61.b)
See problem (3.60.c) for full detail on this compute. The idea and computations are identical.
$$ \cp{A_2}{A_1^c}=\frac{\pr{A_2A_1^c}}{\pr{A_1^c}}=\frac{\cp{A_2A_1^c}{C}\pr{C}+\cp{A_2A_1^c}{C^c}\pr{C^c}}{1-\pr{A_1}} $$
$$ =\frac{\cp{A_2}{C}\cp{A_1^c}{C}\pr{C}+\cp{A_2}{C^c}\cp{A_1^c}{C^c}\pr{C^c}}{1-\frac16} \tag{Independence} $$
$$ =\frac{\frac14\frac34\frac23+0\wts1\wts\frac13}{\frac56}=\frac{\frac6{48}}{\frac56}=\frac18\frac65=\frac3{20} $$
(3.62.a)
Let
- $B$ denote the event that Barbara hit the target
- $D$ denote the event that Dianne hit the target
- $p_b=\pr{B}$
- $p_d=\pr{D}$
- $q_b=\pr{B^c}=1-p_b$
- $q_d=\pr{D^c}=1-p_d$
Note that
$$ \prt{target hit}=1-\prt{target not hit}=1-\prt{neither hit target} $$
$$ =1-\pr{B^cD^c}=1-\pr{B^c}\pr{D^c}=1-q_bq_d $$
Hence
$$ \cpt{both hit target}{target hit}=\frac{\pr{\text{both hit target}\cap\text{target hit}}}{\prt{target hit}} $$
$$ =\frac{\prt{both hit target}}{1-q_bq_d}=\frac{\pr{BD}}{1-q_bq_d}=\frac{\pr{B}\pr{D}}{1-q_bq_d}=\frac{p_bp_d}{1-q_bq_d} $$
(3.62.b)
Note that
$$ \settx{target hit}=\set{(BD),(BD^c),(B^cD)} $$
$$ \settx{Barb hit target}=\set{(BD),(BD^c)} $$
Hence
$$ \cpt{Barb hit target}{target hit}=\frac{\pr{\text{Barb hit target}\cap\text{target hit}}}{\prt{target hit}} $$
$$ =\frac{\prt{Barb hit target}}{1-q_bq_d}=\frac{p_b}{1-q_bq_d} $$