Example 4a
$P(E)=0.5$
$P(F)=0.4$
$P(EF)=0.3$
$P(E\cup F)=P(E)+P(F)-P(EF)=0.5+0.4-0.3=0.6$
The event where J likes neither book is $(E\cup F)^c$ because if $x\in E\cup F$ then $x$ is an event where J likes the first book, or the second book, or both books. Hence if $y\in(E\cup F)^c$, then $y$ is an event where none of those cases are true.
$P\bop(E\cup F)^c\bcp=1-P(E\cup F)=1-0.6=0.4$
Inclusion-Exclusion Identity
I think it is easier to grasp this identity when it is written out. I have written it out for $n=5$. Notice that the first sum has $\binom51=5$ terms, the second sum has $\binom52=10$ terms, the third sum has $\binom53=10$ terms, the fourth sum has $\binom54=5$ terms, and the fifth sum has $\binom55=1$ term. Also notice the alternating signs.
$$ P\Bop\bigcup_{i=1}^{5}E_i\Bcp=\sum_{i=1}^{5}P(E_i)-\sum_{i_1<i_2}P(E_{i_1}E_{i_2})+\sum_{i_1<i_2<i_3}P(E_{i_1}E_{i_2}E_{i_3}) $$
$$ -\sum_{i_1<i_2<i_3<i_4}P(E_{i_1}E_{i_2}E_{i_3}E_{i_4})+P(E_{i_1}E_{i_2}E_{i_3}E_{i_4}E_{i_5}) $$
$$ =P(E_1)+P(E_2)+P(E_3)+P(E_4)+P(E_5) $$
$$ -\Bosb P(E_1E_2)+P(E_1E_3)+P(E_1E_4)+P(E_1E_5)+P(E_2E_3) $$
$$ +P(E_2E_4)+P(E_2E_5)+P(E_3E_4)+P(E_3E_5)+P(E_4E_5)\Bcsb $$
$$ +\Bosb P(E_1E_2E_3)+P(E_1E_2E_4)+P(E_1E_2E_5)+P(E_1E_3E_4)+P(E_1E_3E_5) $$
$$ +P(E_1E_4E_5)+P(E_2E_3E_4)+P(E_2E_3E_5)+P(E_2E_4E_5)+P(E_3E_4E_5)\Bcsb $$
$$ -\Bosb P(E_1E_2E_3E_4)+P(E_1E_2E_3E_5)+P(E_1E_2E_4E_5)+P(E_1E_3E_4E_5)+P(E_2E_3E_4E_5)\Bcsb $$
$$ +P(E_{1}E_{2}E_{3}E_{4}E_{5}) $$
$$ =\sum_{r=1}^{5}(-1)^{r+1}\sum_{i_1<\dots<i_r}P(E_{i_1}\dots E_{i_r}) $$
Example 5b
The number ways that we can get 1 white and 2 black balls with permutation dups removed:
$\binom{6}{1}\binom{5}{2}=6\wt\frac{5\wt4}{2}=60$
The total number of possible picks with permutation dups removed is $\frac{11\wt10\wt9}{3!}=\binom{11}{3}=11\wt5\wt3=165$
So the probability is $\frac{60}{165}=\frac{4\wt15}{11\wt15}=\frac{4}{11}\approx36.4\%$.
Example 5c
All possible committees: $\frac{15\wt14\wt13\wt12\wt11}{5!}=\binom{15}{5}$
Probability of choosing 3 men and 2 women: $\frac{\binom{6}{3}\binom{9}{2}}{\binom{15}{5}}$
Example 5d
All possible k-ball selections with permutation dups removed: $\frac{n\wt(n-1)\wt\dots\wt(n-k)}{k!}=\binom{n}{k}$
Count where special ball is included in k-selection: $\binom{1}{1}\binom{n-1}{k-1}=\binom{n-1}{k-1}=\frac{(n-1)\wt\dots\wt(n-1-(k-1))}{(k-1)!}=\frac{(n-1)\wt\dots\wt(n-k)}{(k-1)!}$
So the probability of picking the special ball with k picks is $\frac{\binom{n-1}{k-1}}{\binom{n}{k}}=\frac{(n-1)\wt\dots\wt(n-k)}{(k-1)!}\frac{k!}{n\wt(n-1)\wt\dots\wt(n-k)}=\frac{k}{n}$
Example 5f
Total number of all possible poker hands, without permutation dups: $\binom{52}{5}=\frac{52\wt51\wt50\wt49\wt48}{5!}$
Total number of straight hands, without permutation dups: $\cancel{52\wt4\wt4\wt4\wt3=9{,}984}$
Total number of straight hands, without permutation dups: Let’s look at one example $A,2,3,4,5$. There are $4^5$ such hands. This count excludes permutation dups because we’ve only considered one card in each slot (i.e. an ace in slot 1, a duece in slot 2, etc). But we do need to remove the straight flush. In the count $4^5$, there are $4$ hands which are straight flushes. So our count for Ace-to-5 is $4^5-4$. And we can apply this same count upto 10, Jack, Queen, King, Ace. So we have a total count of $10(4^5-4)$.
So our probability is $\frac{10(4^5-4)}{\binom{52}{5}}$
Example 5g
Total number of all possible poker hands, without permutation dups: $\binom{52}{5}=\frac{52\wt51\wt50\wt49\wt48}{5!}$
Total number of full house hands, without permutations dups: $\binom{4}{3}\binom{4}{2}\wt13\wt12$
Example 5h
Total $\binom{52}{13,13,13,13}=\frac{52!}{\bop(13)!\bcp^4}$
All spades
$$ 1\wts\binom{39}{13,13,13}+\binom{39}{13}\wts1\wts\binom{26}{13,13}+\binom{39}{13,13}\wts1\wts\binom{13}{13}+\binom{39}{13,13,13}\wts1 $$
$$ =\frac{(39)!}{\bop(13)!\bcp^3}+\frac{(39)!(26)!}{(26)!\bop(13)!\bcp^3}+\frac{(39)!}{\bop(13)!\bcp^3}+\frac{(39)!}{\bop(13)!\bcp^3} $$
$$ =\frac{(39)!}{\bop(13)!\bcp^3}+\frac{(39)!}{\bop(13)!\bcp^3}+\frac{(39)!}{\bop(13)!\bcp^3}+\frac{(39)!}{\bop(13)!\bcp^3} $$
$$ =\frac{4\wts(39)!}{\bop(13)!\bcp^3} $$
Probability
$$ \frac{\bop(13)!\bcp^4}{(52)!}\frac{4\wts(39)!}{\bop(13)!\bcp^3} $$
$$ =\frac{4\wts(13)!(39!)}{(52)!} $$
$$ =\frac{4}{\binom{52}{13}} \tag{matches textbook answer} $$
$$ =\frac{4\wts(13)!}{52\wts51\wts50\wts49\wts48\wts47\wts46\wts45\wts44\wts43\wts42\wts41\wts40} $$
$$ =\frac{52\wts(12)!}{52\wts51\wts50\wts49\wts48\wts47\wts46\wts45\wts44\wts43\wts42\wts41\wts40} $$
$$ =\frac{(12)!}{51\wts50\wts49\wts48\wts47\wts46\wts45\wts44\wts43\wts42\wts41\wts40} $$
In [1034]: d=51*50*49*48*47*46*45*44*43*42*41*40
In [1035]: w=wf(12)
In [1036]: w/d
Out[1036]: 6.299078089796431e-12
Example 5h Part (b)
One ace each count
$$ \binom41\binom{48}{12}\binom31\binom{36}{12}\binom21\binom{24}{12} $$
$$ 4!\binom{48}{12,12,12,12} \tag{matches textbook answer} $$
$$ =\frac{4!\wts48\wts47\wt\wt\wt14\wts13}{\bop(12)!\bcp^3} $$
Probability
$$ =\frac{\bop(13)!\bcp^4}{(52)!}\frac{4!\wts48\wts47\wt\wt\wt14\wts13}{\bop(12)!\bcp^3} $$
$$ =\frac{(13)!13^34!\wts48\wts47\wt\wt\wt14\wts13}{(52)!} $$
$$ =\frac{(13)!13^34!}{52\wts51\wts50\wts49\wts(12)!} $$
$$ =\frac{13^44!}{52\wts51\wts50\wts49} $$
In [1038]: pw(13,4)*wf(4)/(52*51*50*49)
Out[1038]: 0.10549819927971188
Example 5i
Let $n=3$. Then the total number of possible outcomes is $365^3$.
How to compute the number of possible ways that no two of them share the same b-day? Arbitrarily order the persons $p_1$, $p_2$, and $p_3$. Then $p_1$ can have any b-day but $p_2$ must not have the same b-day as $p_1$. So far we have $365\wt364$ possibilities. Similarly, $p_3$ cannot have the same b-day as either of those two. So there are $363$ possible b-days for $p_3$ and we have a total of $365\wt364\wt363$. So the probability that no two of the three shares a birthday is
$$ \frac{365\wts364\wts363}{365^3} $$
Similar for arbitrary $n$
$$ \frac{365\wts364\wts363\wt\wt\wt(365-n+1)}{365^n} $$
Example 5k
Total:
If we just line up the players in a row and pair them off right down the line without regard to order or permutations, then there are $40!$ possible pairings. But we must remove the intra-pairing permutations so we divide by $2^{20}$. We must also remove the inter-pairing permutations so we divide by $20!$:
$$ \frac{40!}{2^{20}20!} $$
Another way to compute this is the standard grouping. First we disregard inter-pairing permutations and compute $\binom{40}{2,2,2,\wt\wt\wt,2,2,2}=\frac{40!}{2^{20}}$. Then we divide by $20!$ to remove the inter-pairing permutations to get the same result.
Zero Pairings Count:
First we recognize that this must mean there are two sub-lines in the line of 40. The first 20 players are all offensive and the second 20 players are all defensive.
To first pair off the 20 offensive players, we do the same thing as above:
$$ \frac{20!}{2^{10}10!} $$
We do the same computation for the defensive players.
Probability:
$$ \frac{\frac{20!}{2^{10}10!}\frac{20!}{2^{10}10!}}{\frac{40!}{2^{20}20!}} $$
$$ =\frac{20!}{2^{10}10!}\frac{20!}{2^{10}10!}\frac{2^{20}20!}{40!} $$
$$ =\frac{(20!)^3}{(10!)^240!} $$
Continuity Proposition 6.1
Proposition: If $\set{E_n,n\geq1}$ is a increasing sequence and we define $\set{F_n,n\geq1}$ as
$$ F_1=E_1 $$
$$ F_n=E_n\Bop\bigcup_1^{n-1}E_i\Bcp^c=E_nE_{n-1}^c \tag{6.1.1} $$
then the $F_n$ are mutually exclusive and
$$ \bigcup_1^n F_i=\bigcup_1^n E_i \tag{6.1.2} $$
Proof:
First we prove the second equality in (6.1.1) by showing that $\bigcup_1^{n-1}E_i=E_{n-1}$. Suppose $x\in\bigcup_1^{n-1}E_i$. Then $x\in E_j$ for some $1\leq j\leq n-1$. But $\set{E_i}$ is an increasing sequence hence $x\in E_j\subset E_{j+1}\subset\dots\subset E_{n-1}$.
Conversely suppose $x\in E_{n-1}$. Then $x\in E_{n-1}\subset\bigcup_1^{n-1}E_i$ and we’re done.
Now suppose that $x\in\bigcup_1^n F_i$. Then $x\in F_j$ for some $1\leq j\leq n$. If $j=1$ then $x\in F_1=E_1\subset \bigcup_1^n E_i$. If $j>1$ then $x\in F_j=E_jE_{j-1}^c\subset E_j\subset \bigcup_1^n E_i$.
Conversely suppose $x\in\bigcup_1^n E_i$. Then $x\in E_j$ for some $1\leq j\leq n$. If $j=1$ then $x\in E_1=F_1\subset \bigcup_1^n F_i$. Suppose instead that $1<j\leq n$. Since $\set{E_i}$ is an increasing sequence, define $j$ to be the smallest index such that $x\in E_j$ so that $x\notin E_i$ for all $i<j$ and $x\in E_i$ for all $i\geq j$. Then $x\in E_jE_{j-1}^c=F_j\subset \bigcup_1^n F_i$.
To show that $F_iF_j=\empty$ for $i<j$, note that $\set{E_n}$ increasing implies $E_iE_j=E_i$ and $\set{E_n^c}$ decreasing implies $E_{j-1}^c\subset E_{i-1}^c$ and $E_{i-1}^cE_{j-1}^c=E_{j-1}^c$. Hence
$$ F_iF_j=E_iE_{i-1}^cE_jE_{j-1}^c=E_iE_jE_{i-1}^cE_{j-1}^c=E_iE_{j-1}^c=\empty $$
since
$$ i<j\implies i\leq j-1\implies E_i\subset E_{j-1} $$
$\wes$
Proposition: If $\set{E_n,n\geq1}$ is a decreasing sequence, then
$$ \bigcup_1^{\infty}E_i^c=\Bop\bigcap_1^{\infty}E_i\Bcp^c $$
Proof:
Let $x\in\bigcup_1^{\infty}E_i^c$. Then $x\in E_j^c$ for some $1\leq j<\infty$. Since $\set{E_n^c}$ is an increasing sequence, define $j$ to be the smallest index such that $x\in E_j^c$ so that $x\notin E_i^c$ for all $i<j$ and $x\in E_i^c$ for all $i\geq j$.
This is equivalent to $x\in E_i$ for all $i<j$ and $x\notin E_i$ for all $i\geq j$.
Hence $x\notin\bigcap_1^{\infty}E_i$ and we’re done.
Conversely, let $x\in\Bop\bigcap_1^{\infty}E_i\Bcp^c$ so that $x\notin\bigcap_1^{\infty}E_i$. Hence $x\notin E_j$ for some $1\leq j<\infty$. Hence $x\in E_j^c$ for some $1\leq j<\infty$ and we’re done.
$\wes$