Conditional Probability and Bayes' Theorem: Very Difficult Problems with Solutions

Solution:
All [tex]\binom{52}{8}[/tex] hands are equally likely, and a hand with exactly [tex]k[/tex] hearts can be chosen in [tex]\binom{13}{k}\binom{39}{8-k}[/tex] ways. Let [tex]A[/tex] be the event “all [tex]8[/tex] cards are hearts” and [tex]B[/tex] the event “at least [tex]3[/tex] cards are hearts”. The hands favorable to [tex]B[/tex] are easiest to count through the complementary event ([tex]0[/tex], [tex]1[/tex] or [tex]2[/tex] hearts):
[tex]|B|=\binom{52}{8}-\binom{39}{8}-13\binom{39}{7}-\binom{13}{2}\binom{39}{6}[/tex]
Since [tex]A\subset B[/tex], we have [tex]A\cap B=A[/tex] and
[tex]P(A\,|\,B)=\frac{P(A\cap B)}{P(B)}=\frac{|A|}{|B|}=\frac{\binom{13}{8}}{\binom{52}{8}-\binom{39}{8}-13\binom{39}{7}-\binom{13}{2}\binom{39}{6}}[/tex]
Numerically [tex]|A|=1287[/tex] and [tex]|B|=236\,577\,627[/tex], so the probability equals [tex]\frac{1}{183\,821}[/tex]. It is wrong to “fix” three hearts and choose the other [tex]5[/tex] cards from the remaining [tex]49[/tex], which gives [tex]\frac{\binom{10}{5}}{\binom{49}{5}}[/tex]: the condition does not say which three cards are hearts, and such counting does not give equally likely hands.

Solution:
Let [tex]H_1[/tex] be the hypothesis “the chosen batch has no defective parts” and [tex]H_2[/tex] the hypothesis “in the chosen batch [tex]\frac{1}{4}[/tex] of the parts are defective”; [tex]P(H_1)=P(H_2)=\frac{1}{2}[/tex]. Let [tex]A[/tex] be the event “the first part is good”. Then [tex]P(A\,|\,H_1)=1[/tex], [tex]P(A\,|\,H_2)=\frac{3}{4}[/tex], and by the total probability formula
[tex]P(A)=\frac{1}{2}\times 1+\frac{1}{2}\times \frac{3}{4}=\frac{7}{8}[/tex]
By Bayes' formula, after the first draw the probabilities of the hypotheses become
[tex]P(H_1\,|\,A)=\frac{\frac{1}{2}\times 1}{\frac{7}{8}}=\frac{4}{7},\quad P(H_2\,|\,A)=\frac{\frac{1}{2}\times \frac{3}{4}}{\frac{7}{8}}=\frac{3}{7}[/tex]
The first part was returned, so for a given batch the second draw does not depend on the first: the second part is defective with probability [tex]0[/tex] in the first batch and with probability [tex]\frac{1}{4}[/tex] in the second. By the total probability formula with the updated probabilities of the hypotheses, the required probability is
[tex]\frac{4}{7}\times 0+\frac{3}{7}\times \frac{1}{4}=\frac{3}{28}[/tex]

Solution:
Same-sex twins are twice as likely as twins of different sexes, and the two probabilities add up to [tex]1[/tex], so the twins are of different sexes with probability [tex]\frac{1}{3}[/tex]. In that case both orders of birth are equally likely, so the probability [tex]r[/tex] that the first twin is a boy and the second a girl is
[tex]r=\frac{1}{2}\times \frac{1}{3}=\frac{1}{6}[/tex]
Let [tex]p[/tex] be the probability that both twins are boys. The first twin is a boy either when both are boys or when the first is a boy and the second a girl, so
[tex]p+r=0.51\Rightarrow p=\frac{51}{100}-\frac{1}{6}=\frac{153-50}{300}=\frac{103}{300}[/tex]
By the definition of conditional probability, the probability that the second twin is a boy, given that the first one is a boy, is
[tex]\frac{p}{p+r}=\frac{\frac{103}{300}}{\frac{153}{300}}=\frac{103}{153}\approx 0.673[/tex]

Solution:
We use the total probability formula. Let [tex]H_k[/tex] ([tex]k=0, 1, 2, 3[/tex]) be the hypothesis that exactly [tex]k[/tex] new balls were taken for the first game. All [tex]\binom{15}{3}=455[/tex] triples are equally likely, so [tex]P(H_k)=\frac{\binom{9}{k}\binom{6}{3-k}}{455}[/tex]:
[tex]P(H_0)=\frac{20}{455},\quad P(H_1)=\frac{135}{455},\quad P(H_2)=\frac{216}{455},\quad P(H_3)=\frac{84}{455}[/tex]
After the first game the box contains [tex]9-k[/tex] new balls, so for the event [tex]A[/tex] “three new balls are taken for the second game” we have [tex]P(A\,|\,H_k)=\frac{\binom{9-k}{3}}{455}[/tex]:
[tex]P(A\,|\,H_0)=\frac{84}{455},\quad P(A\,|\,H_1)=\frac{56}{455},\quad P(A\,|\,H_2)=\frac{35}{455},\quad P(A\,|\,H_3)=\frac{20}{455}[/tex]
Therefore
[tex]P(A)=\sum_{k=0}^{3}P(H_k)\times P(A\,|\,H_k)=\frac{20\times 84+135\times 56+216\times 35+84\times 20}{455^2}[/tex]
[tex]=\frac{18480}{207025}=\frac{528}{5915}\approx 0.089[/tex]

Solution:
All [tex]\binom{5}{3}=10[/tex] triples of balls from the first urn are equally likely. Three black balls cannot be drawn, so there are three hypotheses:
[tex]H_1[/tex]: three white balls are drawn, [tex]P(H_1)=\frac{1}{10}[/tex]; three white balls are moved and the second urn contains [tex]5[/tex] white and [tex]3[/tex] black balls;
[tex]H_2[/tex]: two white balls and one black ball are drawn, [tex]P(H_2)=\frac{\binom{3}{2}\binom{2}{1}}{10}=\frac{6}{10}[/tex]; two white balls are moved and the second urn contains [tex]4[/tex] white and [tex]3[/tex] black balls;
[tex]H_3[/tex]: one white ball and two black balls are drawn, [tex]P(H_3)=\frac{\binom{3}{1}\binom{2}{2}}{10}=\frac{3}{10}[/tex]; two black balls are moved and the second urn contains [tex]2[/tex] white and [tex]5[/tex] black balls.
Let [tex]A[/tex] be the event that both balls drawn from the second urn are white. Then
[tex]P(A\,|\,H_1)=\frac{\binom{5}{2}}{\binom{8}{2}}=\frac{10}{28}=\frac{5}{14}[/tex]
[tex]P(A\,|\,H_2)=\frac{\binom{4}{2}}{\binom{7}{2}}=\frac{6}{21}=\frac{2}{7}[/tex]
[tex]P(A\,|\,H_3)=\frac{\binom{2}{2}}{\binom{7}{2}}=\frac{1}{21}[/tex]
By the total probability formula
[tex]P(A)=\frac{1}{10}\times \frac{5}{14}+\frac{6}{10}\times \frac{2}{7}+\frac{3}{10}\times \frac{1}{21}=\frac{5}{140}+\frac{24}{140}+\frac{2}{140}=\frac{31}{140}[/tex]
Two black balls were moved exactly when [tex]H_3[/tex] occurred, so by Bayes' formula
[tex]P(H_3\,|\,A)=\frac{P(H_3)\,P(A\,|\,H_3)}{P(A)}=\frac{\frac{2}{140}}{\frac{31}{140}}=\frac{2}{31}[/tex]

Solution:
Each player independently passes the information on correctly with probability [tex]\frac{1}{3}[/tex] and reverses it with probability [tex]\frac{2}{3}[/tex]. The announced information is correct exactly when the number of reversals is even. Let [tex]T[/tex] be the event "A told the truth" and [tex]R[/tex] the event "D announced the correct information".
Let [tex]q_0[/tex] and [tex]q_1[/tex] be the probabilities that players B, C and D together lie an even, respectively an odd, number of times:
[tex]q_0=\left(\frac{1}{3}\right)^3+3\times \frac{1}{3}\times \left(\frac{2}{3}\right)^2=\frac{1}{27}+\frac{12}{27}=\frac{13}{27}[/tex]
[tex]q_1=3\times \left(\frac{1}{3}\right)^2\times \frac{2}{3}+\left(\frac{2}{3}\right)^3=\frac{6}{27}+\frac{8}{27}=\frac{14}{27}[/tex]
If A told the truth, the announcement is correct when B, C and D lie an even number of times; if A lied, when they lie an odd number of times. By independence
[tex]P(T\cap R)=\frac{1}{3}\times \frac{13}{27}=\frac{13}{81},\qquad P(\overline{T}\cap R)=\frac{2}{3}\times \frac{14}{27}=\frac{28}{81}[/tex]
Hence [tex]P(R)=\frac{13}{81}+\frac{28}{81}=\frac{41}{81}[/tex], and by Bayes' formula
[tex]P(T\,|\,R)=\frac{P(T\cap R)}{P(R)}=\frac{\frac{13}{81}}{\frac{41}{81}}=\frac{13}{41}[/tex]

Solution:
Let urn [tex]i[/tex] ([tex]i=1 , 2 , 3[/tex]) contain [tex]n_i\ge 1[/tex] balls, [tex]b_i[/tex] of them black, where [tex]b_1+b_2+b_3=6[/tex]. Each urn is chosen with probability [tex]\frac{1}{3}[/tex], and a white ball is drawn from urn [tex]i[/tex] with probability [tex]1-\frac{b_i}{n_i}[/tex]. By the total probability formula
[tex]P=\frac{1}{3}\left(3-\frac{b_1}{n_1}-\frac{b_2}{n_2}-\frac{b_3}{n_3}\right)=1-\frac{1}{3}\left(\frac{b_1}{n_1}+\frac{b_2}{n_2}+\frac{b_3}{n_3}\right)[/tex]
The other two urns contain at least one ball each, so [tex]n_i\le 18-2=16[/tex] and [tex]\frac{b_i}{n_i}\ge\frac{b_i}{16}[/tex] for every [tex]i[/tex]. Hence
[tex]\frac{b_1}{n_1}+\frac{b_2}{n_2}+\frac{b_3}{n_3}\ge\frac{b_1+b_2+b_3}{16}=\frac{6}{16}=\frac{3}{8}\Rightarrow P\le 1-\frac{1}{3}\times \frac{3}{8}=\frac{7}{8}[/tex]
Equality is reached when all [tex]6[/tex] black balls are put into one urn together with [tex]10[/tex] white balls, and each of the other two urns contains one white ball:
[tex]P=\frac{1}{3}\left(1+1+\frac{10}{16}\right)=\frac{7}{8}[/tex]
So the greatest probability is [tex]\frac{7}{8}[/tex]. (The original arrangement gives only [tex]\frac{1}{3}\left(\frac{5}{6}+\frac{3}{6}+\frac{4}{6}\right)=\frac{2}{3}[/tex].)

Solution:
Number the balls. All [tex](n+m+k)![/tex] orders in which they can be drawn are equally likely, and the event depends only on which of the [tex]n+m[/tex] white and red balls comes first among them.
For any two white or red balls [tex]X[/tex] and [tex]Y[/tex], exchanging the places of [tex]X[/tex] and [tex]Y[/tex] in every order is a bijection between the orders in which [tex]X[/tex] is the first of the white and red balls and those in which [tex]Y[/tex] is the first of them. Hence each of the [tex]n+m[/tex] white and red balls is the first among them with the same probability [tex]\frac{1}{n+m}[/tex]. Exactly [tex]n[/tex] of these balls are white, so the required probability is [tex]\frac{n}{n+m}[/tex].

Solution:
Let [tex]N=n+m+k[/tex]. The draws are independent, and in each of them a white, red or green ball appears with probability [tex]\frac{n}{N}[/tex], [tex]\frac{m}{N}[/tex], [tex]\frac{k}{N}[/tex] respectively. A white ball comes before a red one exactly when, for some [tex]j\ge 0[/tex], the first [tex]j[/tex] draws give green balls and draw number [tex]j+1[/tex] gives a white ball. These events are mutually exclusive, and by independence the one with index [tex]j[/tex] has probability [tex]\left(\frac{k}{N}\right)^j\frac{n}{N}[/tex]. Since [tex]0\le\frac{k}{N}\lt 1[/tex], summing the geometric series gives
[tex]P=\sum_{j=0}^{\infty}\left(\frac{k}{N}\right)^j\frac{n}{N}=\frac{n}{N}\times \frac{1}{1-\frac{k}{N}}=\frac{n}{N-k}=\frac{n}{n+m}[/tex]
So the required probability equals [tex]\frac{n}{n+m}[/tex]; the green balls do not influence it.

Solution:
Let [tex]S=M+N[/tex]; the total amount of money never changes. Let [tex]p_n[/tex] be the probability that John wins when he has [tex]n[/tex] dollars (and Pierre has [tex]S-n[/tex]). Clearly [tex]p_0=0[/tex] and [tex]p_S=1[/tex].
For [tex]0\lt n\lt S[/tex] consider the first toss: with probability [tex]\frac{1}{2}[/tex] John's money becomes [tex]n+1[/tex] and with probability [tex]\frac{1}{2}[/tex] it becomes [tex]n-1[/tex], and since the tosses are independent, the game then continues as if it started anew. By the total probability formula
[tex]p_n=\frac{1}{2}p_{n+1}+\frac{1}{2}p_{n-1}\Rightarrow p_{n+1}-p_n=p_n-p_{n-1}[/tex]
So [tex]p_0 , p_1 , \ldots , p_S[/tex] is an arithmetic progression that goes from [tex]0[/tex] to [tex]1[/tex] in [tex]S[/tex] equal steps, hence [tex]p_n=\frac{n}{S}[/tex]. John starts with [tex]M[/tex] dollars, so
[tex]P=p_M=\frac{M}{M+N}[/tex]
In the same way Pierre wins with probability [tex]\frac{N}{M+N}[/tex]; the two probabilities add up to [tex]1[/tex], so the game ends with probability [tex]1[/tex].

Solution:
Let [tex]A_k[/tex] be the event that the [tex]k[/tex]-th ball drawn is white. If the first [tex]k-1[/tex] balls were white, before the [tex]k[/tex]-th draw the urn contains [tex]1+2(k-1)=2k-1[/tex] white balls and [tex]1[/tex] black ball, so [tex]P(A_k\,|\,A_1\cap\ldots\cap A_{k-1})=\frac{2k-1}{2k}[/tex]. By the multiplication rule the required probability is
[tex]P=P(A_1\cap A_2\cap\ldots\cap A_{50})=\frac{1}{2}\times \frac{3}{4}\times \frac{5}{6}\times \ldots\times \frac{99}{100}[/tex]
Consider also
[tex]Q=\frac{2}{3}\times \frac{4}{5}\times \frac{6}{7}\times \ldots\times \frac{100}{101}[/tex]
For every [tex]k[/tex] we have [tex]\frac{2k-1}{2k}\lt\frac{2k}{2k+1}[/tex], because this is equivalent to [tex](2k-1)(2k+1)=4k^2-1\lt 4k^2[/tex]. Hence each factor of [tex]P[/tex] is smaller than the corresponding factor of [tex]Q[/tex], so [tex]P\lt Q[/tex] and
[tex]P^2\lt P\times Q=\frac{1\times 2\times 3\times \ldots\times 100}{2\times 3\times 4\times \ldots\times 101}=\frac{1}{101}\lt\frac{1}{100}[/tex]
Therefore [tex]P\lt\frac{1}{10}=0.1[/tex]. (In fact [tex]P=\frac{\binom{100}{50}}{2^{100}}\approx 0.0796[/tex].)

Solution:
Let [tex]A_i[/tex] be the event that the [tex]i[/tex]-th pair drawn consists of a boy and a girl, [tex]i=1 , 2 , \ldots , N[/tex]. By the multiplication rule
[tex]P(A_1\cap\ldots\cap A_N)=P(A_1)\times P(A_2\,|\,A_1)\times \ldots\times P(A_N\,|\,A_1\cap\ldots\cap A_{N-1})[/tex]
If the first [tex]i-1[/tex] pairs were "boy – girl", the box contains [tex]N-i+1[/tex] boys and [tex]M-i+1[/tex] girls, i.e. [tex]N+M-2i+2[/tex] cards. All [tex]\binom{N+M-2i+2}{2}[/tex] pairs of them are equally likely, and [tex](N-i+1)(M-i+1)[/tex] of them are "boy – girl". Hence
[tex]P(A_i\,|\,A_1\cap\ldots\cap A_{i-1})=\frac{(N-i+1)(M-i+1)}{\binom{N+M-2i+2}{2}}=\frac{2(N-i+1)(M-i+1)}{(N+M-2i+2)(N+M-2i+1)}[/tex]
Multiply these for [tex]i=1 , 2 , \ldots , N[/tex]. The numerators give
[tex]2^N\times N!\times M(M-1)\times \ldots\times (M-N+1)=\frac{2^N\times N!\times M!}{(M-N)!}[/tex]
and the denominators give the product of all integers from [tex]M-N+1[/tex] to [tex]N+M[/tex], i.e. [tex]\frac{(N+M)!}{(M-N)!}[/tex]. Therefore
[tex]P=\frac{2^N\times N!\times M!}{(M-N)!}\times \frac{(M-N)!}{(N+M)!}=\frac{2^N\times N!\times M!}{(N+M)!}[/tex]

Solution:
Let passenger [tex]i[/tex] have seat [tex]i[/tex]. The first passenger takes each of the seats [tex]2 , 3 , \ldots , 100[/tex] with probability [tex]\frac{1}{99}[/tex]. If he takes seat [tex]100[/tex], the last passenger certainly fails.
Suppose he takes seat [tex]k[/tex] with [tex]2\le k\le 99[/tex] (probability [tex]\frac{98}{99}[/tex]). Passengers [tex]2 , \ldots , k-1[/tex] take their own seats and passenger [tex]k[/tex] chooses a random free seat. From then on, until one of the seats [tex]1[/tex] and [tex]100[/tex] is taken, every passenger who finds his seat taken chooses at random among free seats that include both seat [tex]1[/tex] and seat [tex]100[/tex], so at each such choice these two seats are equally likely.
If seat [tex]1[/tex] is taken first, the occupied seats become exactly the seats of the passengers already on board, so everyone after that finds his own seat free and the last passenger gets seat [tex]100[/tex]. If seat [tex]100[/tex] is taken first, the last passenger fails. When the last passenger boards, only one seat is free, so one of the seats [tex]1[/tex] and [tex]100[/tex] has been taken before; by symmetry each of them is taken first with probability [tex]\frac{1}{2}[/tex]. Therefore
[tex]P=\frac{98}{99}\times \frac{1}{2}=\frac{49}{99}[/tex]
(For [tex]n[/tex] seats the answer is [tex]\frac{n-2}{2(n-1)}[/tex]; the well-known answer [tex]\frac{1}{2}[/tex] is for a first passenger who may also choose his own seat.)

Solution:
All [tex]n![/tex] orders in which the cards can come out are equally likely. Only the relative order of the numbers matters, so we may assume that the numbers are [tex]1 , 2 , \ldots , n[/tex] and the outcome is a random permutation [tex]\left(x_1 , x_2 , \ldots , x_n\right)[/tex].
For each [tex]k[/tex] let [tex]R_k[/tex] be the rank of [tex]x_k[/tex] among the first [tex]k[/tex] numbers, i.e. the number of indices [tex]i\le k[/tex] with [tex]x_i\le x_k[/tex]. Then [tex]1\le R_k\le k[/tex], and [tex]A_k[/tex] occurs exactly when [tex]R_k=k[/tex].
The permutation can be recovered from its ranks: [tex]x_n[/tex] is the [tex]R_n[/tex]-th smallest of all [tex]n[/tex] numbers; after removing it, [tex]x_{n-1}[/tex] is the [tex]R_{n-1}[/tex]-th smallest of the remaining numbers, and so on. So the map [tex]\left(x_1 , \ldots , x_n\right)\mapsto\left(R_1 , \ldots , R_n\right)[/tex] is injective, and its values lie in the set [tex]T[/tex] of all integer sequences [tex]\left(r_1 , \ldots , r_n\right)[/tex] with [tex]1\le r_k\le k[/tex] for every [tex]k[/tex].
The set [tex]T[/tex] has [tex]1\times 2\times \ldots\times n=n![/tex] elements, as many as there are permutations, so the map is a bijection and every element of [tex]T[/tex] occurs with probability [tex]\frac{1}{n!}[/tex]. The sequences in [tex]T[/tex] with [tex]r_k=k[/tex] are obtained by fixing the [tex]k[/tex]-th coordinate and choosing the others freely, so there are [tex]\frac{n!}{k}[/tex] of them. Therefore
[tex]P(A_k)=\frac{n!}{k}\times \frac{1}{n!}=\frac{1}{k}[/tex]
In the same way, for any indices [tex]1\le i_1\lt i_2\lt\ldots\lt i_s\le n[/tex] the sequences in [tex]T[/tex] with [tex]r_{i_1}=i_1 , \ldots , r_{i_s}=i_s[/tex] are [tex]\frac{n!}{i_1\times i_2\times \ldots\times i_s}[/tex] in number, so
[tex]P(A_{i_1}\cap A_{i_2}\cap\ldots\cap A_{i_s})=\frac{1}{i_1\times i_2\times \ldots\times i_s}=P(A_{i_1})\times P(A_{i_2})\times \ldots\times P(A_{i_s})[/tex]
This product rule holds for every subfamily of the events, so the events [tex]A_1 , A_2 , \ldots , A_n[/tex] are independent.