Conditional Probability and Bayes' Theorem: Problems with Solutions

Solution:
Let [tex]A[/tex] be the event “the card is a six” and [tex]B[/tex] the event “the value of the card is greater than [tex]3[/tex] and less than [tex]8[/tex]”. The event [tex]B[/tex] consists of the fours, fives, sixes and sevens, i.e. of [tex]4\times 4=16[/tex] cards, and [tex]A\cap B[/tex] consists of the [tex]4[/tex] sixes. All [tex]52[/tex] cards are equally likely, so by the definition of conditional probability
[tex]P(A\,|\,B)=\frac{P(A\cap B)}{P(B)}=\frac{4/52}{16/52}=\frac{4}{16}=\frac{1}{4}[/tex]
In other words, the condition reduces the sample space to [tex]16[/tex] equally likely cards, [tex]4[/tex] of which are sixes.

Solution:
The [tex]36[/tex] ordered outcomes [tex]\left(a , b\right)[/tex] are equally likely. Let [tex]B[/tex] be the event “the sum is [tex]8[/tex]” and [tex]A[/tex] the event “exactly one six”. The event [tex]B[/tex] consists of [tex]5[/tex] outcomes (the outcome [tex]\left(4 , 4\right)[/tex] is counted once):
[tex]\left(2 , 6\right),\ \left(3 , 5\right),\ \left(4 , 4\right),\ \left(5 , 3\right),\ \left(6 , 2\right)\ \Rightarrow\ P(B)=\frac{5}{36}[/tex]
The event [tex]A\cap B[/tex] consists of the outcomes [tex]\left(2 , 6\right)[/tex] and [tex]\left(6 , 2\right)[/tex], so [tex]P(A\cap B)=\frac{2}{36}[/tex]. By the definition of conditional probability
[tex]P(A\,|\,B)=\frac{P(A\cap B)}{P(B)}=\frac{2/36}{5/36}=\frac{2}{5}[/tex]

Solution:
Let [tex]R[/tex] be the event “the business offers a pension plan” and [tex]L[/tex] the event “the business offers life insurance”. Then
[tex]P(R)=0.42,\quad P(L)=0.61,\quad P(R\cap L)=0.32[/tex]
By the definition of conditional probability
[tex]P(R\,|\,L)=\frac{P(R\cap L)}{P(L)}=\frac{0.32}{0.61}=\frac{32}{61}\approx 0.52[/tex]
Note that the reverse conditional probability is different: [tex]P(L\,|\,R)=\frac{0.32}{0.42}=\frac{16}{21}[/tex].

Solution:
All [tex]200[/tex] cars are equally likely. Let [tex]A[/tex] be the event “the car is of make N” and [tex]B[/tex] the event “the car is yellow”. There are [tex]2+3+5=10[/tex] yellow cars, and [tex]5[/tex] of them are of make N:
[tex]P(B)=\frac{10}{200},\quad P(A\cap B)=\frac{5}{200}[/tex]
Therefore
[tex]P(A\,|\,B)=\frac{P(A\cap B)}{P(B)}=\frac{5/200}{10/200}=\frac{1}{2}[/tex]
Equivalently, the information that the car is yellow reduces the sample space to the [tex]10[/tex] yellow cars, and [tex]5[/tex] of them are of make N.

Solution:
Let [tex]A_1[/tex] be the event that the first attempt is correct and [tex]A_2[/tex] the event that the second attempt is correct. All [tex]10[/tex] digits are equally likely, so [tex]P(A_1)=\frac{1}{10}[/tex]. If the first attempt fails (probability [tex]\frac{9}{10}[/tex]), Ivan chooses among the [tex]9[/tex] remaining digits, so [tex]P(A_2\,|\,\overline{A_1})=\frac{1}{9}[/tex].
The events [tex]A_1[/tex] and [tex]\overline{A_1}\cap A_2[/tex] are mutually exclusive, hence
[tex]P=P(A_1)+P(\overline{A_1})\times P(A_2\,|\,\overline{A_1})=\frac{1}{10}+\frac{9}{10}\times \frac{1}{9}=\frac{2}{10}=\frac{1}{5}[/tex]
Equivalently: Ivan tries [tex]2[/tex] of the [tex]10[/tex] digits, and the correct digit is among them with probability [tex]\frac{2}{10}[/tex].

Solution:
Let [tex]A[/tex] be the event “at least one machine needs attention”. Its complementary event [tex]\overline{A}[/tex] is “none of the four machines needs attention”. The machines work independently, so
[tex]P(\overline{A})=0.9\times 0.8\times 0.75\times 0.7=0.378[/tex]
Therefore
[tex]P(A)=1-P(\overline{A})=1-0.378=0.622[/tex]

Solution:
All [tex]10[/tex] machines produce the same number of parts, so a random part comes from a machine of the first, second or third type (hypotheses [tex]H_1[/tex], [tex]H_2[/tex], [tex]H_3[/tex]) with probabilities
[tex]P(H_1)=\frac{5}{10}=0.5,\quad P(H_2)=\frac{3}{10}=0.3,\quad P(H_3)=\frac{2}{10}=0.2[/tex]
These hypotheses are mutually exclusive and cover all cases. Let [tex]A[/tex] be the event “the part is good”; then [tex]P(A\,|\,H_1)=0.94[/tex], [tex]P(A\,|\,H_2)=0.9[/tex] and [tex]P(A\,|\,H_3)=0.85[/tex]. By the total probability formula
[tex]P(A)=0.5\times 0.94+0.3\times 0.9+0.2\times 0.85=0.47+0.27+0.17=0.91[/tex]

Solution:
The second refrigerator contains [tex]30-12=18[/tex] cartons. Let [tex]H_1[/tex] and [tex]H_2[/tex] be the events that the first or the second refrigerator is chosen, and [tex]A[/tex] the event “the carton contains apple juice”. Then
[tex]P(H_1)=P(H_2)=\frac{1}{2},\quad P(A\,|\,H_1)=\frac{3}{12}=\frac{1}{4},\quad P(A\,|\,H_2)=\frac{7}{18}[/tex]
By the total probability formula
[tex]P(A)=P(H_1)P(A\,|\,H_1)+P(H_2)P(A\,|\,H_2)=\frac{1}{2}\times \frac{1}{4}+\frac{1}{2}\times \frac{7}{18}=\frac{9}{72}+\frac{14}{72}=\frac{23}{72}[/tex]
Note that the answer is not [tex]\frac{10}{30}[/tex]: the [tex]30[/tex] cartons are not equally likely to be chosen (each carton in the first refrigerator is chosen with probability [tex]\frac{1}{24}[/tex], and each carton in the second one with probability [tex]\frac{1}{36}[/tex]).

Solution:
Let [tex]p[/tex] be the probability that the first shooter hits; the second shooter misses with probability [tex]1-0.8=0.2[/tex]. Exactly one hit means either “the first hits and the second misses” or “the first misses and the second hits”. These two events are mutually exclusive, and the shots are independent, so we multiply the probabilities within each case and add the cases:
[tex]p\times 0.2+(1-p)\times 0.8=0.38[/tex]
Hence
[tex]0.8-0.6p=0.38\ \Rightarrow\ 0.6p=0.42\ \Rightarrow\ p=0.7[/tex]

Solution:
Let [tex]O_1[/tex] be the event "the first candy is orange" and [tex]O_2[/tex] the event "the second candy is orange". All [tex]n[/tex] candies are equally likely to be taken first, and [tex]6[/tex] of them are orange:
[tex]P(O_1)=\frac{6}{n}[/tex]
If the first candy is orange, [tex]n-1[/tex] candies remain and [tex]5[/tex] of them are orange, so
[tex]P(O_2\,|\,O_1)=\frac{5}{n-1}[/tex]
By the multiplication rule, the probability that both candies are orange is
[tex]P(O_1\cap O_2)=P(O_1)\times P(O_2\,|\,O_1)=\frac{6}{n}\times \frac{5}{n-1}=\frac{30}{n(n-1)}[/tex]
By the condition this probability equals [tex]\frac{1}{3}[/tex]:
[tex]\frac{30}{n(n-1)}=\frac{1}{3}\iff n(n-1)=90\iff n^2-n-90=0[/tex]
which had to be proved. (The roots of this equation are [tex]n=10[/tex] and [tex]n=-9[/tex], so the bag contained [tex]10[/tex] candies.)

Solution:
Let [tex]A[/tex] be the event "the first ball is white" and [tex]B[/tex] the event "the two balls have different colors". By the multiplication rule, the probability that the first ball is white and the second is red, and the probability that the first ball is red and the second is white, are
[tex]\frac{5}{16}\times \frac{11}{15}=\frac{11}{48},\qquad \frac{11}{16}\times \frac{5}{15}=\frac{11}{48}[/tex]
Event [tex]B[/tex] occurs in exactly one of these two ways, and [tex]A\cap B[/tex] is the first of them:
[tex]P(B)=\frac{11}{48}+\frac{11}{48}=\frac{11}{24},\qquad P(A\cap B)=\frac{11}{48}[/tex]
By the definition of conditional probability
[tex]P(A\,|\,B)=\frac{P(A\cap B)}{P(B)}=\frac{11}{48}\times \frac{24}{11}=\frac{1}{2}[/tex]
(The two orders, white then red and red then white, are equally likely, which explains the answer.)

Solution:
All [tex]220[/tex] fans are equally likely to be chosen, so every conditional probability is a ratio of counts:
[tex]P(X\,|\,Y)=\frac{|X\cap Y|}{|Y|}[/tex]
From [tex]P(L\,|\,W)=\frac{1}{2}[/tex], half of the [tex]110[/tex] women, i.e. [tex]55[/tex], support Levski, so the number of women supporting CSKA is
[tex]|W\cap C|=110-55=55[/tex]
Let [tex]c=|C|[/tex] be the number of CSKA fans. Since [tex]P(M\,|\,C)=\frac{8}{19}[/tex], the women make up [tex]1-\frac{8}{19}=\frac{11}{19}[/tex] of the CSKA fans:
[tex]\frac{11}{19}\,c=55\Rightarrow c=95[/tex]
Hence the number of men supporting CSKA is
[tex]|M\cap C|=95-55=40[/tex]

Solution:
Consider the hypotheses [tex]H_1[/tex]: "a die is chosen" and [tex]H_2[/tex]: "the tetrahedron is chosen", and the event [tex]A[/tex]: "[tex]3[/tex] dots come up". Then
[tex]P(H_1)=\frac{4}{5},\quad P(H_2)=\frac{1}{5},\quad P(A\,|\,H_1)=\frac{1}{6},\quad P(A\,|\,H_2)=\frac{1}{4}[/tex]
By the total probability formula
[tex]P(A)=\frac{4}{5}\times \frac{1}{6}+\frac{1}{5}\times \frac{1}{4}=\frac{2}{15}+\frac{1}{20}=\frac{11}{60}[/tex]
By Bayes' formula
[tex]P(H_1\,|\,A)=\frac{P(H_1)\times P(A\,|\,H_1)}{P(A)}=\frac{2}{15}\times \frac{60}{11}=\frac{8}{11}[/tex]

Solution:
Consider the hypotheses [tex]M[/tex]: "the person is a man" and [tex]W[/tex]: "the person is a woman", and the event [tex]D[/tex]: "the person is color-blind". Then
[tex]P(M)=P(W)=\frac{1}{2},\quad P(D\,|\,M)=0.05,\quad P(D\,|\,W)=0.0025[/tex]
By the total probability formula
[tex]P(D)=\frac{1}{2}\times 0.05+\frac{1}{2}\times 0.0025=0.025+0.00125=0.02625[/tex]
By Bayes' formula
[tex]P(M\,|\,D)=\frac{P(M)\times P(D\,|\,M)}{P(D)}=\frac{0.025}{0.02625}=\frac{2500}{2625}=\frac{20}{21}[/tex]