Conditional Probability and Bayes' Theorem: Difficult Problems with Solutions

Solution:
The two answers are chosen independently, and each is correct with probability [tex]\frac{1}{4}[/tex]. Let [tex]A[/tex] be the event "both answers are correct" and [tex]B[/tex] the event "at least one answer is correct". The complementary event of [tex]B[/tex] is "both answers are wrong", so
[tex]P(A)=\frac{1}{4}\times \frac{1}{4}=\frac{1}{16},\qquad P(B)=1-\frac{3}{4}\times \frac{3}{4}=\frac{7}{16}[/tex]
Since [tex]A\subset B[/tex], we have [tex]P(A\cap B)=P(A)[/tex], and by the definition of conditional probability
[tex]P(A\,|\,B)=\frac{P(A\cap B)}{P(B)}=\frac{1}{16}\times \frac{16}{7}=\frac{1}{7}[/tex]
(Equivalently: of the [tex]16[/tex] equally likely pairs of answers, [tex]7[/tex] contain a correct answer, and only [tex]1[/tex] of them has both answers correct.)

Solution:
The equally likely outcomes are the [tex]\binom{52}{2}=1326[/tex] unordered pairs of cards. Let [tex]S[/tex] be the event "the two cards have the same rank" and [tex]D[/tex] the event "the two cards are of different suits". Two cards of the same rank always have different suits, so [tex]S\cap D=S[/tex]. Count the pairs of the same rank and the pairs of the same suit:
[tex]|S|=13\times \binom{4}{2}=78,\qquad 4\times \binom{13}{2}=312[/tex]
Hence event [tex]D[/tex] consists of [tex]1326-312=1014[/tex] pairs and
[tex]P(S\,|\,D)=\frac{P(S\cap D)}{P(D)}=\frac{78}{1014}=\frac{1}{13}[/tex]
Shorter reasoning: whatever the first card is, under the condition the second card is equally likely to be any of the [tex]39[/tex] cards of the other three suits, and exactly [tex]3[/tex] of them have the same rank as the first card, so [tex]P=\frac{3}{39}=\frac{1}{13}[/tex].

Solution:
Write [tex]\text{H}[/tex] for heads and [tex]\text{T}[/tex] for tails. The [tex]2^3=8[/tex] sequences of results are equally likely. Two events [tex]X[/tex] and [tex]Y[/tex] are independent when [tex]P(X\cap Y)=P(X)\times P(Y)[/tex]. We have [tex]A=\{\text{HHH},\text{HHT},\text{HTH},\text{HTT}\}[/tex], [tex]B=\{\text{HHH},\text{HHT},\text{HTH},\text{THH}\}[/tex] and [tex]C=\{\text{HHH},\text{TTT}\}[/tex], so
[tex]P(A)=\frac{4}{8}=\frac{1}{2},\quad P(B)=\frac{4}{8}=\frac{1}{2},\quad P(C)=\frac{2}{8}=\frac{1}{4}[/tex]
[tex]A\cap B=\{\text{HHH},\text{HHT},\text{HTH}\}[/tex], so [tex]P(A\cap B)=\frac{3}{8}\ne\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}[/tex]: the events [tex]A[/tex] and [tex]B[/tex] are dependent.
[tex]A\cap C=\{\text{HHH}\}[/tex], so [tex]P(A\cap C)=\frac{1}{8}=\frac{1}{2}\times \frac{1}{4}[/tex]: the events [tex]A[/tex] and [tex]C[/tex] are independent.
[tex]B\cap C=\{\text{HHH}\}[/tex], so [tex]P(B\cap C)=\frac{1}{8}=\frac{1}{2}\times \frac{1}{4}[/tex]: the events [tex]B[/tex] and [tex]C[/tex] are independent.
Hence exactly [tex]2[/tex] of the three pairs are independent.

Solution:
There are [tex]23[/tex] balls. If the first ball is white, [tex]12[/tex] of the remaining [tex]22[/tex] balls are white; if it is red, [tex]13[/tex] of the remaining [tex]22[/tex] balls are white:
[tex]P(A_1)=\frac{13}{23},\quad P(\overline{A_1})=\frac{10}{23},\quad P(A_2\,|\,A_1)=\frac{12}{22},\quad P(A_2\,|\,\overline{A_1})=\frac{13}{22}[/tex]
By the multiplication rule and the total probability formula
[tex]P(A_1\cap A_2)=\frac{13}{23}\times \frac{12}{22}[/tex]
[tex]P(A_2)=\frac{13}{23}\times \frac{12}{22}+\frac{10}{23}\times \frac{13}{22}=\frac{13\times (12+10)}{23\times 22}=\frac{13}{23}[/tex]
So the second ball is white with the same probability as the first. Therefore
[tex]P(A_1\,|\,A_2)=\frac{P(A_1\cap A_2)}{P(A_2)}=\frac{13}{23}\times \frac{12}{22}\times \frac{23}{13}=\frac{12}{22}=\frac{6}{11}[/tex]

Solution:
Let [tex]H_1[/tex], [tex]H_2[/tex], [tex]H_3[/tex] be the events "box [tex]1[/tex], [tex]2[/tex], [tex]3[/tex] is chosen", respectively, and let [tex]N[/tex] be the event "the drawn ball is new". The box is chosen at random, and once the box is fixed, the probability of a new ball is the share of new balls in it:
[tex]P(H_1)=P(H_2)=P(H_3)=\frac{1}{3}, \quad P(N\,|\,H_1)=\frac{2}{3}, \quad P(N\,|\,H_2)=\frac{3}{5}, \quad P(N\,|\,H_3)=\frac{4}{6}=\frac{2}{3}[/tex]
By the total probability formula
[tex]P(N)=\frac{1}{3}\left(\frac{2}{3}+\frac{3}{5}+\frac{2}{3}\right)=\frac{1}{3}\times \frac{29}{15}=\frac{29}{45}[/tex]
By Bayes' formula
[tex]P(H_1\,|\,N)=\frac{P(H_1)\,P(N\,|\,H_1)}{P(N)}=\frac{\frac{1}{3}\times \frac{2}{3}}{\frac{29}{45}}=\frac{2}{9}\times \frac{45}{29}=\frac{10}{29}[/tex]
The answer is not [tex]\frac{2}{9}[/tex] (two of the nine new balls), because balls in different boxes are not equally likely to be drawn: a particular ball in box [tex]1[/tex] is drawn with probability [tex]\frac{1}{3}\times \frac{1}{3}=\frac{1}{9}[/tex], while a particular ball in box [tex]2[/tex] is drawn with probability [tex]\frac{1}{3}\times \frac{1}{5}=\frac{1}{15}[/tex].

Solution:
Consider a randomly chosen question. Let [tex]K[/tex] be the event "the student knows the answer", [tex]G[/tex] the event "the student guesses" and [tex]C[/tex] the event "the marked answer is correct". A guess picks each of the four answers with equal probability, so
[tex]P(K)=\frac{2}{3}, \quad P(G)=\frac{1}{3}, \quad P(C\,|\,K)=1, \quad P(C\,|\,G)=\frac{1}{4}[/tex]
By the total probability formula
[tex]P(C)=\frac{2}{3}\times 1+\frac{1}{3}\times \frac{1}{4}=\frac{2}{3}+\frac{1}{12}=\frac{3}{4}[/tex]
The answer was not picked at random exactly when the student knew it, so by Bayes' formula
[tex]P(K\,|\,C)=\frac{P(K)\,P(C\,|\,K)}{P(C)}=\frac{\frac{2}{3}}{\frac{3}{4}}=\frac{8}{9}[/tex]

Solution:
Let [tex]D[/tex] be the event "the person is ill" and [tex]T[/tex] the event "the test is positive". From the conditions
[tex]P(D)=0.005, \quad P(\overline{D})=0.995, \quad P(T\,|\,D)=0.99, \quad P(T\,|\,\overline{D})=1-0.99=0.01[/tex]
By the total probability formula
[tex]P(T)=P(D)\,P(T\,|\,D)+P(\overline{D})\,P(T\,|\,\overline{D})[/tex]
[tex]P(T)=0.005\times 0.99+0.995\times 0.01=0.00495+0.00995=0.0149[/tex]
By Bayes' formula
[tex]P(D\,|\,T)=\frac{P(D)\,P(T\,|\,D)}{P(T)}=\frac{0.00495}{0.0149}=\frac{99}{298}\approx 0.33[/tex]
Intuitively: among [tex]200\,000[/tex] people, on average [tex]1000[/tex] are ill and [tex]990[/tex] of them test positive, while [tex]1990[/tex] of the [tex]199\,000[/tex] healthy people also test positive. So only [tex]990[/tex] of the [tex]2980[/tex] positive results belong to ill people.

Solution:
Consider the hypotheses [tex]H_1[/tex]: "the coin with two heads is chosen", [tex]H_2[/tex]: "the ordinary coin is chosen" and [tex]H_3[/tex]: "the coin with two tails is chosen". The coin is chosen at random, so [tex]P(H_1)=P(H_2)=P(H_3)=\frac{1}{3}[/tex]. Let [tex]A[/tex] be the event "the coin shows heads". Then
[tex]P(A\,|\,H_1)=1, \quad P(A\,|\,H_2)=\frac{1}{2}, \quad P(A\,|\,H_3)=0[/tex]
By the total probability formula
[tex]P(A)=\frac{1}{3}\left(1+\frac{1}{2}+0\right)=\frac{1}{2}[/tex]
The hidden side is heads exactly when the chosen coin is the one with two heads, so by Bayes' formula the required probability is
[tex]P(H_1\,|\,A)=\frac{P(H_1)\,P(A\,|\,H_1)}{P(A)}=\frac{\frac{1}{3}\times 1}{\frac{1}{2}}=\frac{2}{3}[/tex]
Intuitively: three equally likely heads faces can end up on top, and two of them belong to the coin with two heads.

Solution:
Let [tex]H_1[/tex] be the event "the lost ball is white", [tex]H_2[/tex] the event "the lost ball is black" and [tex]A[/tex] the event "the drawn ball is white". After a white ball is lost, the urn contains [tex]3[/tex] white and [tex]3[/tex] black balls; after a black ball is lost, it contains [tex]4[/tex] white and [tex]2[/tex] black balls. Hence
[tex]P(H_1)=\frac{4}{7}, \quad P(H_2)=\frac{3}{7}, \quad P(A\,|\,H_1)=\frac{3}{6}=\frac{1}{2}, \quad P(A\,|\,H_2)=\frac{4}{6}=\frac{2}{3}[/tex]
By the total probability formula
[tex]P(A)=\frac{4}{7}\times \frac{1}{2}+\frac{3}{7}\times \frac{2}{3}=\frac{2}{7}+\frac{2}{7}=\frac{4}{7}[/tex]
This is to be expected: losing a ball unseen does not change the probability that a randomly drawn ball is white. By Bayes' formula
[tex]P(H_1\,|\,A)=\frac{P(H_1)\,P(A\,|\,H_1)}{P(A)}=\frac{\frac{2}{7}}{\frac{4}{7}}=\frac{1}{2}[/tex]

Solution:
Let [tex]H_k[/tex] be the event "exactly [tex]k[/tex] white balls were moved", [tex]k=0, 1, 2, 3[/tex]; these hypotheses are mutually exclusive and cover all cases. All [tex]\binom{9}{3}=84[/tex] choices of [tex]3[/tex] balls are equally likely, so [tex]P(H_k)=\frac{\binom{4}{k}\binom{5}{3-k}}{84}[/tex]:
[tex]P(H_0)=\frac{10}{84}, \quad P(H_1)=\frac{40}{84}, \quad P(H_2)=\frac{30}{84}, \quad P(H_3)=\frac{4}{84}[/tex]
Let [tex]A[/tex] be the event "the ball drawn from the second vessel is white". If [tex]H_k[/tex] occurred, the second vessel holds [tex]k[/tex] white balls out of [tex]3[/tex], so [tex]P(A\,|\,H_k)=\frac{k}{3}[/tex]. By the total probability formula
[tex]P(A)=\frac{10}{84}\times 0+\frac{40}{84}\times \frac{1}{3}+\frac{30}{84}\times \frac{2}{3}+\frac{4}{84}\times 1=\frac{40+60+12}{252}=\frac{112}{252}=\frac{4}{9}[/tex]
This equals the probability of drawing a white ball directly from the first vessel, as it should: the ball drawn at the end is equally likely to be any of the [tex]9[/tex] balls.

Solution:
The shots are independent, so the probability of each combination of hits and misses is the product of the individual probabilities. Exactly two hits occur in three mutually exclusive ways:
[tex]A[/tex] and [tex]B[/tex] hit, [tex]C[/tex] misses: [tex]0.6\times 0.5\times 0.6=0.18[/tex];
[tex]A[/tex] and [tex]C[/tex] hit, [tex]B[/tex] misses: [tex]0.6\times 0.5\times 0.4=0.12[/tex];
[tex]B[/tex] and [tex]C[/tex] hit, [tex]A[/tex] misses: [tex]0.4\times 0.5\times 0.4=0.08[/tex].
Hence the probability of exactly two hits is [tex]0.18+0.12+0.08=0.38[/tex], and the probability that there are exactly two hits and [tex]C[/tex] is one of the hitters is [tex]0.12+0.08=0.20[/tex]. By the definition of conditional probability, the required probability is
[tex]P=\frac{0.20}{0.38}=\frac{20}{38}=\frac{10}{19}[/tex]
Since [tex]\frac{10}{19}\gt\frac{1}{2}[/tex], it is more likely that [tex]C[/tex] hit than that he missed, although [tex]C[/tex] is the weakest shooter.

Solution:
Let [tex]H_k[/tex] be the event "exactly [tex]k[/tex] units are broken", [tex]k=0, 1, 2, 3[/tex]. The units break down independently, each with probability [tex]\frac{1}{4}[/tex], so by the Bernoulli formula [tex]P(H_k)=\binom{3}{k}\left(\frac{1}{4}\right)^k\left(\frac{3}{4}\right)^{3-k}[/tex]:
[tex]P(H_0)=\frac{27}{64}, \quad P(H_1)=\frac{27}{64}, \quad P(H_2)=\frac{9}{64}, \quad P(H_3)=\frac{1}{64}[/tex]
Let [tex]A[/tex] be the event "the machine stops". By the conditions of the problem
[tex]P(A\,|\,H_0)=0, \quad P(A\,|\,H_1)=\frac{2}{3}, \quad P(A\,|\,H_2)=\frac{3}{4}, \quad P(A\,|\,H_3)=1[/tex]
By the total probability formula
[tex]P(A)=\frac{27}{64}\times \frac{2}{3}+\frac{9}{64}\times \frac{3}{4}+\frac{1}{64}\times 1=\frac{72}{256}+\frac{27}{256}+\frac{4}{256}=\frac{103}{256}[/tex]

Solution:
The questions are drawn without replacement. The first two questions form one of [tex]\binom{25}{2}=300[/tex] equally likely pairs. Let [tex]B_0[/tex] be the event "the student knows neither of them" and [tex]B_1[/tex] the event "he knows exactly one of them". There are [tex]\binom{10}{2}=45[/tex] pairs of unknown questions and [tex]15\times 10=150[/tex] pairs with exactly one known question, so
[tex]P(B_0)=\frac{45}{300}=\frac{3}{20}, \quad P(B_1)=\frac{150}{300}=\frac{1}{2}[/tex]
If [tex]B_1[/tex] occurred, the remaining [tex]23[/tex] questions contain [tex]14[/tex] known and [tex]9[/tex] unknown ones, so the third question is unknown with conditional probability [tex]\frac{9}{23}[/tex]. The student fails if [tex]B_0[/tex] occurs, or if [tex]B_1[/tex] occurs and he does not know the third question. These cases are mutually exclusive, so the required probability is
[tex]P=P(B_0)+P(B_1)\times \frac{9}{23}=\frac{3}{20}+\frac{1}{2}\times \frac{9}{23}=\frac{3}{20}+\frac{9}{46}=\frac{69+90}{460}=\frac{159}{460}\approx 0.346[/tex]

Solution:
All throws are independent. Consider one round: a throw of [tex]A[/tex], then of [tex]B[/tex], then of [tex]C[/tex]. In this round [tex]A[/tex] wins with probability [tex]p_A[/tex]; [tex]B[/tex] wins if [tex]A[/tex] fails and [tex]B[/tex] throws [tex]4[/tex] or [tex]5[/tex] (probability [tex]p_B[/tex]); [tex]C[/tex] wins if [tex]A[/tex] and [tex]B[/tex] fail and [tex]C[/tex] throws [tex]6[/tex] (probability [tex]p_C[/tex]); nobody wins with probability [tex]q[/tex]:
[tex]p_A=\frac{3}{6}=\frac{1}{2},\quad p_B=\frac{1}{2}\times \frac{2}{6}=\frac{1}{6}[/tex]
[tex]p_C=\frac{1}{2}\times \frac{4}{6}\times \frac{1}{6}=\frac{1}{18},\quad q=\frac{1}{2}\times \frac{4}{6}\times \frac{5}{6}=\frac{5}{18}[/tex]
If nobody wins, the game starts over in exactly the same situation, so [tex]C[/tex] wins in the [tex]k[/tex]-th round with probability [tex]q^{k-1}p_C[/tex]. Summing the geometric series:
[tex]P(C)=p_C\left(1+q+q^2+\ldots\right)=\frac{p_C}{1-q}=\frac{\frac{1}{18}}{1-\frac{5}{18}}=\frac{1}{13}[/tex]
Check: in the same way [tex]P(A)=\frac{p_A}{1-q}=\frac{9}{13}[/tex] and [tex]P(B)=\frac{p_B}{1-q}=\frac{3}{13}[/tex], and indeed [tex]\frac{9}{13}+\frac{3}{13}+\frac{1}{13}=1[/tex].

Solution:
The [tex]36[/tex] outcomes of the two dice are equally likely, and the lifted hand is chosen independently of the dice, each hand with probability [tex]\frac{1}{2}[/tex]. Let [tex]S[/tex] be the event “the shown die is a [tex]6[/tex]” and [tex]H[/tex] the event “both dice show [tex]6[/tex]”. We use the total probability formula with three hypotheses: two sixes (probability [tex]\frac{1}{36}[/tex], then [tex]S[/tex] is certain), exactly one six ([tex]10[/tex] outcomes, probability [tex]\frac{10}{36}[/tex], then [tex]S[/tex] occurs with probability [tex]\frac{1}{2}[/tex]) and no six (then [tex]S[/tex] is impossible):
[tex]P(S)=\frac{1}{36}\times 1+\frac{10}{36}\times \frac{1}{2}+\frac{25}{36}\times 0=\frac{6}{36}=\frac{1}{6}[/tex]
By Bayes' formula
[tex]P(H\,|\,S)=\frac{P(H)\times P(S\,|\,H)}{P(S)}=\frac{\frac{1}{36}\times 1}{\frac{1}{6}}=\frac{1}{6}[/tex]
This agrees with the intuition that the hidden die does not depend on the shown one. The answer [tex]\frac{1}{11}[/tex] would be correct only if we merely knew that at least one of the dice shows [tex]6[/tex]; here we know more, because a randomly chosen die was shown.

Solution:
By symmetry we may assume that you first chose door [tex]1[/tex]. With probability [tex]\frac{1}{3}[/tex] the prize is behind door [tex]1[/tex], and then switching loses. With probability [tex]\frac{2}{3}[/tex] the prize is behind door [tex]2[/tex] or door [tex]3[/tex]; then the host is forced to open the other, empty door, so the remaining closed door hides the prize and switching wins. Hence switching wins with probability [tex]\frac{2}{3}[/tex] (staying wins only with probability [tex]\frac{1}{3}[/tex]).
The same result by Bayes' formula: suppose the host opened door [tex]3[/tex]; let [tex]C_i[/tex] be the event “the prize is behind door [tex]i[/tex]” and [tex]D[/tex] the event “the host opens door [tex]3[/tex]”. Then [tex]P(D\,|\,C_1)=\frac{1}{2}[/tex], [tex]P(D\,|\,C_2)=1[/tex], [tex]P(D\,|\,C_3)=0[/tex], so
[tex]P(C_2\,|\,D)=\frac{\frac{1}{3}\times 1}{\frac{1}{3}\times \frac{1}{2}+\frac{1}{3}\times 1+\frac{1}{3}\times 0}=\frac{2}{3}[/tex]

Solution:
Let [tex]A_k[/tex] be the event that the [tex]k[/tex]-th person gets the coin, [tex]1\le k\le 10[/tex]. It happens exactly when the first [tex]k-1[/tex] people take pieces without the coin and the [tex]k[/tex]-th person takes the piece with the coin. We use the multiplication rule for successive draws without replacement.
The first person misses the coin with probability [tex]\frac{9}{10}[/tex]; if he missed it, [tex]9[/tex] pieces remain, one of them with the coin, so the second person misses it with probability [tex]\frac{8}{9}[/tex], and so on. If the first [tex]k-1[/tex] people missed the coin, [tex]11-k[/tex] pieces remain and exactly one of them contains it, so the [tex]k[/tex]-th person gets it with conditional probability [tex]\frac{1}{11-k}[/tex]. Therefore
[tex]P(A_k)=\frac{9}{10}\times \frac{8}{9}\times \ldots\times \frac{11-k}{12-k}\times \frac{1}{11-k}=\frac{1}{10}[/tex]
because the product telescopes (for [tex]k=1[/tex] it is simply [tex]\frac{1}{10}[/tex]). So all ten people get the coin with the same probability [tex]\frac{1}{10}[/tex]: a person who draws later has a larger conditional chance [tex]\frac{1}{11-k}[/tex], but this is exactly compensated by the smaller chance that the coin has not been taken yet.
Another argument: the order in which the [tex]10[/tex] pieces are taken is a random permutation of the pieces, and the piece with the coin is equally likely to stand in each of the [tex]10[/tex] places.

Solution:
Recall that the events [tex]A_1, \ldots, A_n[/tex] are independent if for every [tex]k\ge 2[/tex] and every choice of indices [tex]1\le i_1\lt i_2\lt\ldots\lt i_k\le n[/tex]
[tex]P(A_{i_1}\cap A_{i_2}\cap\ldots\cap A_{i_k})=P(A_{i_1})\times P(A_{i_2})\times \ldots\times P(A_{i_k})[/tex]
Lemma: if [tex]A_1, A_2, \ldots, A_n[/tex] are independent, then [tex]\overline{A_1}, A_2, \ldots, A_n[/tex] are also independent. Indeed, the equalities that do not involve the first event do not change. For an equality that involves it, take any non-empty set of indices [tex]S\subseteq\{2, \ldots, n\}[/tex] and let [tex]B=\bigcap_{i\in S}A_i[/tex]. The events [tex]A_1\cap B[/tex] and [tex]\overline{A_1}\cap B[/tex] are disjoint and their union is [tex]B[/tex], so, using the independence of the original events twice, we get
[tex]P(\overline{A_1}\cap B)=P(B)-P(A_1\cap B)=\prod_{i\in S}P(A_i)-P(A_1)\prod_{i\in S}P(A_i)[/tex]
[tex]=\left(1-P(A_1)\right)\prod_{i\in S}P(A_i)=P(\overline{A_1})\prod_{i\in S}P(A_i)[/tex]
which is the required equality. The definition of independence is symmetric in the events, so the lemma allows us to replace any single event of an independent family by its complement.
Now we prove by induction on [tex]k=0, 1, \ldots, n[/tex] that the events [tex]\overline{A_1}, \ldots, \overline{A_k}, A_{k+1}, \ldots, A_n[/tex] are independent. Base case: for [tex]k=0[/tex] this is the hypothesis. Inductive step: if the statement holds for some [tex]k\lt n[/tex], applying the lemma to the event [tex]A_{k+1}[/tex] gives it for [tex]k+1[/tex]. For [tex]k=n[/tex] we obtain that [tex]\overline{A_1}, \overline{A_2}, \ldots, \overline{A_n}[/tex] are independent.