Conditional Probability: Definitions, Formulas, and Solved Problems

In this case, the selection of a female student occurs after the condition that a student from class B has been selected. Therefore, this is a case of conditional probability.
Let $E$ denote the event that a female student is selected. Let $F$ denote the event that a student from class B is selected. This means that $n(F) = 3 + 4 = 7$ and $n(E \cap F) = 3.$ Thus, we obtain $$p(E \mid F) = \dfrac{n(E \cap F)}{n(F)} = \dfrac{3}{7}.$$Therefore, the probability that a female student is selected under these conditions is $\boxed{\dfrac37}$
(Answer B)

Let $X$ and $Y$ respectively denote the events that a fifth-semester student is selected and that a student receives grade A in the Introduction to Chemistry course.
In this case, we seek the value of $p(X \mid Y).$
Since the number of fifth-semester students who received grade A is $n(X \cap Y) = 3$ and the number of students who received grade A is $n(Y) = 3+10+5=18,$ we obtain
$$\begin{aligned} p(X \mid Y) & = \dfrac{n(X \cap Y)}{n(Y)} \\ & = \dfrac{3}{18} \\ & = \dfrac16. \end{aligned}$$Therefore, the probability that the selected student is a fifth-semester student given that the student received grade A is $\boxed{\dfrac16}$
(Answer C)

In this experiment, the occurrence of the first die showing $1$ happens after the condition that the sum of the dice faces is less than $4.$ Therefore, this is a case of conditional probability.
Let $E$ denote the event that the first die shows $1.$ Let $F$ denote the event that the sum of the dice faces is less than $4.$ This means that $$F = \{(1, 1), (1, 2), (2, 1)\}$$and $$E \cap F = \{(1, 1), (1, 2)\}$$so that $n(F) = 3$ and $n(E \cap F) = 2.$ Thus, we obtain $$p(E \mid F) = \dfrac{n(E \cap F)}{n(F)} = \dfrac{2}{3}.$$Therefore, the probability that the first die shows $1$ given that the sum of the dice faces is less than $4$ is $\boxed{\dfrac23}$
(Answer B)

Let $E$ denote the event that the sum of the dice faces is greater than $9.$ Let $F$ denote the event that the first die shows $5,$ namely $\{(5, 1), (5, 2), \cdots, (5, 6)\}.$ Thus, we obtain $E \cap F = \{(5, 5), (5, 6)\}.$ This means that $p(F) = \dfrac{6}{6^2} = \dfrac{1}{6}$ and $p(E \cap F) = \dfrac{2}{6^2} = \dfrac{1}{18}$ so that $$p(E \mid F) = \dfrac{p(E \cap F)}{p(F)} = \dfrac{1/18}{1/6} = \dfrac13.$$Therefore, the probability that the sum of the dice faces is greater than $9$ given that the first die shows $5$ is $\boxed{\dfrac13}$
(Answer B)

Let $E$ denote the event that a defective bulb is selected on the first draw and let $F$ denote the event that a defective bulb is selected on the second draw. The notation $E \cap F$ can be interpreted as the occurrence of event $E,$ followed by the occurrence of $F$ (after $E$ occurs).
The probability of event $E$ is $P(E) = \dfrac{6}{30} = \dfrac15,$ while the probability of event $F$ is $P(F \mid E) = \dfrac{6-1}{30-1} = \dfrac{5}{29}.$
Thus, the probability that the two selected bulbs are defective is $$\begin{aligned} P(E \cap F) & = P(E) \cdot P(F \mid E) \\ & = \dfrac15 \cdot \dfrac{5}{29} \\ & = \dfrac{1}{29}. \end{aligned}$$(Answer C)

Let $E$ denote the event that the player wins a round in the first session. Let $F$ denote the event that the player wins a round in the second session (which means the player must first win a round in the first session). Thus, we obtain $n(E) = 30$ and $n(F \cap E) = 15$ so that $p(F \mid E) = \dfrac{n(F \cap E)}{n(E)} =\dfrac{15}{30}.$
(Answer D)

This case is a conditional probability case because the event that the patient sues the doctor occurs after the event that the doctor incorrectly diagnoses the disease.
Let $E$ denote the event that the patient sues the doctor. Let $F$ denote the event that the doctor incorrectly diagnoses the disease. It is known that the probability that a doctor diagnoses a disease correctly is $p(F^c) = 0,\!75.$ This means that $p(F) = 1-p(F^c) = 1-0,\!75 = 0,\!25$ and $p(E \mid F) = 0,\!92.$ Thus, we obtain
$$\begin{aligned} p(E \mid F) & = \dfrac{p(E \cap F)}{p(F)} \\ 0,\!92 & = \dfrac{p(E \cap F)}{0,\!25} \\ p(E \cap F) & = 0,\!92 \cdot 0,\!25 = 0,\!23. \end{aligned}$$Therefore, the probability that the doctor incorrectly diagnoses the disease and is sued by the patient is $\boxed{0,\!23}$
(Answer B)

Let $E$ denote the event that exactly $4$ number sides appear. Let $F$ denote the event that a number side appears on the first toss. This means that $n(F) = 1 \times 2^4 = 16$ and $n(E \cap F) = 4$ because the number of arrangements for exactly three remaining number sides to appear in the next four tosses is $4.$ Thus, we obtain $p(E \mid F) = \dfrac{n(E \cap F)}{n(F)} = \dfrac{4}{16} = \dfrac14.$
Therefore, the probability of the event that exactly $4$ number sides appear given that the first toss results in a number side is $\boxed{\dfrac14}$
(Answer C)

Let $E$ denote the event that the oldest family member is male. Let $F$ denote the event that the family consists of some males and some females. This means that $n(F) = 2^5-2 = 30,$ obtained from the number of arrangements of $5$ family members based on gender minus the number of events where all family members are male or all are female. In addition, $n(E \cap F) = 2^4-1 = 15,$ obtained from the number of arrangements of the remaining $4$ family members based on gender minus the number of events where the remaining $4$ family members are all male.
Thus, we obtain $p(E \mid F) = \dfrac{n(E \cap F)}{n(F)} = \dfrac{15}{30} = \dfrac12.$ Therefore, the probability that the family consists of some males and some females is $\boxed{\dfrac12}$
(Answer A)

Let $E$ denote the event that exactly $4$ number sides appear. Let $F$ denote the event that a number side appears on the first toss. This means that $n(F) = 1 \times 2^4 = 16$ and $n(E \cap F) = 4$ because the number of arrangements for exactly three remaining number sides to appear in the next four tosses is $4.$ Thus, we obtain $p(E \mid F) = \dfrac{n(E \cap F)}{n(F)} = \dfrac{4}{16} = \dfrac14.$
Therefore, the probability of the event that exactly $4$ number sides appear given that the first toss results in a number side is $\boxed{\dfrac14}$
(Answer C)

Let $E$ denote the event that the oldest family member is male. Let $F$ denote the event that the family consists of some males and some females. This means $n(F) = 2^5-2 = 30,$ obtained from the number of arrangements of $5$ family members based on gender minus the number of events where all family members are male or female. In addition, $n(E \cap F) = 2^4-1 = 15,$ obtained from the number of arrangements of the remaining $4$ family members based on gender minus the number of events when the remaining $4$ family members are all male.
Thus, we obtain $p(E \mid F) = \dfrac{n(E \cap F)}{n(F)} = \dfrac{15}{30} = \dfrac12.$ Therefore, the probability that the family consists of some males and some females is $\boxed{\dfrac12}$
(Answer A)

Let $E$ denote the event that a bitstring of length four containing at least two consecutive $0$ bits is formed. Let $F$ denote the event that a bitstring of length four whose first bit is $0$ is formed. Since $E \cap F = \{0000, 0001, 0010, 0011, 0100\},$ we obtain $p(E \cap F) = \dfrac{5}{16}.$ Since there are $8$ bitstrings of length four whose first bit is $0,$ we obtain $p(F) = \dfrac{8}{16} = \dfrac12.$ Consequently, $p(E \mid F) = \dfrac{5/16}{1/2} = \dfrac58.$
Therefore, the probability of the event that the formed bitstring contains at least two consecutive $0$ bits given that the first bit is $0$ is $\boxed{\dfrac58}$
(Answer C)

Using the definition of conditional probability, we obtain
$$p(A \mid B) = p(B \mid A) \Rightarrow \dfrac{p(A \cap B)}{p(B)}= \dfrac{p(B \cap A)}{p(A)}.$$Since $p(A \cap B) = p(B \cap A),$ we obtain $\dfrac{1}{p(B)} = \dfrac{1}{p(A)}$ which means that $p(A) = p(B).$ In other words, the probabilities of events $A$ and $B$ occurring must be equal for the equation $p(A \mid B) = p(B \mid A)$ to hold.
(Answer A)

Let $B$ and $A$ respectively denote the events that the fire truck and the ambulance are ready to operate when an emergency situation occurs. Since $B$ and $A$ are independent events, the relation $p(B \cap A) = p(B) \cdot p(A)$ applies, where $P(B \cap A)$ denotes the probability that both the fire truck and ambulance are ready to operate. Since $p(B) = 0,\!96$ and $P(A) = 0,\!88,$ we obtain
$$\begin{aligned} p(B \cap A) & = p(B) \cdot p(A) \\ & = 0,\!96 \cdot 0,\!88 \\ & = 0,\!8448. \end{aligned}$$Therefore, the probability that both vehicles are ready to operate is $\boxed{0,\!8448}$
(Answer C)

Two events $A$ and $B$ are said to be mutually exclusive if $p(A \cap B) = 0$ and are said to be independent if $p(A \cap B) = p(A) \cap p(B).$
Suppose the experiment conducted is rolling a six-sided die.
Answer a)
Suppose $A$ is the event of obtaining $2, 3,$ or $5,$ while $B$ is the event of obtaining $7.$ From this, we obtain $p(A) = \dfrac36$ and $p(B) = 0.$ Furthermore, $A \cap B = \{ \}$ so that $p(A \cap B) = 0.$ Therefore, we get $p(A \cap B) = p(A) \cdot p(B) = 0.$ This shows that $A$ and $B$ are mutually exclusive and independent at the same time.
Answer b)
Suppose $A$ is the event of obtaining $2, 3,$ or $5,$ while $B$ is the event of obtaining $4$ or $6.$ From this, we obtain $p(A) = \dfrac36$ and $p(B) = \dfrac26.$ Furthermore, $A \cap B = \{ \}$ so that $p(A \cap B) = 0.$ Therefore, we get $$p(A \cap B) = 0 \ne p(A) \cap p(B) = \dfrac36 \cdot \dfrac26 = \dfrac16.$$This shows that $A$ and $B$ are mutually exclusive, but not independent.
Answer c)
Suppose $A$ is the event of obtaining $1$ or $4,$ while $B$ is the event of obtaining $1, 3$ or $5.$ From this, we obtain $p(A) = \dfrac26$ and $p(B) = \dfrac36.$ Furthermore, $A \cap B = \{1 \}$ so that $p(A \cap B) = \dfrac16.$ Therefore, we get $$p(A \cap B) = \dfrac16 = \dfrac26 \cdot \dfrac36 = p(A) \cdot p(B).$$This shows that $A$ and $B$ are independent, but not mutually exclusive.
Answer d)
Suppose $A$ is the event of obtaining $1$ or $4,$ while $B$ is the event of obtaining $1, 4$ or $5.$ From this, we obtain $p(A) = \dfrac26$ and $p(B) = \dfrac36.$ Furthermore, $A \cap B = \{1, 4 \}$ so that $p(A \cap B) = \dfrac26.$ Therefore, we get $$p(A \cap B) = \dfrac26 \ne \dfrac26 \cdot \dfrac36 = p(A) \cdot p(B).$$This shows that $A$ and $B$ are neither mutually exclusive nor independent.

It is known that $X$ denotes the event that a criminal commits a robbery and $Y$ denotes the event that the criminal successfully escapes.
Answer a)
$p(X \mid Y)$ denotes the probability that the criminal commits a robbery after successfully escaping.
Answer b)
$p(Y^c \mid X)$ denotes the probability that the criminal fails to escape after committing a robbery.
Answer c)
$P(X^c \mid Y^c)$ denotes the probability that the criminal does not commit a robbery after failing to escape.

Let $A, B,$ and $C$ respectively denote the events of selecting a heavy smoker, a person suffering from hypertension, and a non-smoker.
Answer a)
In this case, we will determine the value of $p(B \mid A),$ namely
$$\begin{aligned} p(B \mid A) & = \dfrac{n(B \cap A)}{n(A)} \\ & = \dfrac{30}{30 + 19} \\ & = \dfrac{30}{49}. \end{aligned}$$Therefore, the probability that the person suffers from hypertension given that he or she is a heavy smoker is $\boxed{\dfrac{30}{49}}$
Answer b)
In this case, we will determine the value of $p(C \mid B^c),$ namely
$$\begin{aligned} p(C \mid B^c) & = \dfrac{n(C \cap B^c)}{n(B^c)} \\ & = \dfrac{48}{48 + 26 + 19} \\ & = \dfrac{16}{31}. \end{aligned}$$Therefore, the probability that the person is a non-smoker given that he or she does not suffer from hypertension is $\boxed{\dfrac{16}{31}}$

Let $A$ and $B$ respectively denote the events that the car undergoes an oil change and replaces the oil filter, so that it is known that $p(A) = 0,\!26,$ $p(B) = 0,\!40,$ and $p(A \cap B) = 0,\!14.$
Answer a)
In this case, we will determine the value of $p(B \mid A),$ namely
$$p(B \mid A) = \dfrac{p(B \cap A)}{p(A)} = \dfrac{p(A \cap B)}{p(A)} = \dfrac{0,\!14}{0,\!26} = \dfrac{7}{13}.$$Therefore, the probability that the car replaces the oil filter given that it has undergone an oil change is $\boxed{\dfrac{7}{13}}$
Answer b)
In this case, we will determine the value of $p(A \mid B),$ namely
$$p(A \mid B) = \dfrac{p(A \cap B)}{p(B)} = \dfrac{0,\!14}{0,\!40} = \dfrac{7}{20}.$$Therefore, the probability that the car undergoes an oil change given that it has replaced the oil filter is $\boxed{\dfrac{7}{20}}$

Let $E$ denote the event that the Dragon Air plane departs on time. Let $F$ denote the event that the Dragon Air plane arrives on time. It is known that $p(E) = 0,\!90,$ $p(F) = 0,\!80,$ and $p(E \cap F) = 0,\!75.$
Answer a)
Since the event that the Dragon Air plane arrives on time occurs after the plane departs on time, this is a case of conditional probability.
$$\begin{aligned} p(F \mid E) & = \dfrac{p(E \cap F)}{p(E)} \\ & = \dfrac{0,\!75}{0,\!90} \\ & \approx 0,\!833 \end{aligned}$$Therefore, the probability that the Dragon Air plane arrives on time given that the plane departs on time is $\boxed{0,\!833}$
Answer b)
Observe that $p(E) = 0,\!90,$ $p(F) = 0,\!80,$ and $p(E \cap F) = 0,\!75.$ Since $0,\!90 \cdot 0,\!80 = \!0,72 \neq 0,\!75,$ it is concluded that $p(E) \cdot p(F) \neq p(E \cap F).$ In other words, the events that the Dragon Air plane departs and arrives on time are not independent events.

Suppose $H$ and $A$ respectively represent the events of a rainy day and wind speed exceeding the normal limit. Let $n(S)$ represent the number of days in the $6$-month period so that $n(S) = 180.$ It is known that $n(H) = 150$ and $n(A) = 120$ as well as $n(A \mid H) = 108.$
Answer a)
During the next year’s $6$-month rainy season, the probability of a rainy day is $$p(H) = \dfrac{n(H)}{n(S)} = \dfrac{150}{180} = \dfrac56.$$Then, the probability that the wind speed exceeds the normal limit is $$p(A) = \dfrac{n(A)}{n(S)} = \dfrac{120}{180} = \dfrac23.$$Answer b)
Observe that
$$\begin{aligned} p(A \mid H) & = \dfrac{p(A \cap H)}{p(H)} \\ \dfrac{108}{150} & = \dfrac{p(A \cap H)}{p(H)} \\ p(A \cap H) & = \dfrac35. \end{aligned}$$Thus,
$$p(A) \cdot p(H) = \dfrac23 \cdot \dfrac56 = \dfrac59 \neq p(A \cap H).$$Since $p(A) \cdot p(H) \ne p(A \cap H),$ it is concluded that the rainy day event and the event of wind speed exceeding the normal limit are not independent.
Answer c)
During the next year’s $6$-month rainy season, the probability of a rainy day given that the wind speed exceeds the normal limit is
$$\begin{aligned} p(H \mid A) & = \dfrac{p(H \cap A)}{p(A)} \\ & = \dfrac{3/5}{2/3} \\ & = \dfrac{9}{10}. \end{aligned}$$

There are $8$ binary strings of length $4$ beginning with bit $1.$
$$\begin{array}{cc} \hline 1000 & 1001 \\ 1010 & 1100 \\ 1011 & 1101 \\ 1110 & 1111 \\ \hline \end{array}$$In addition, there are $8$ binary strings of length $4$ containing an even number of bit $1$s.
$$\begin{array}{cc} \hline 1111 & 0000 \\ 1100 & 1010 \\ 1001 & 0110 \\ 0101 & 0011 \\ \hline \end{array}$$Since there are $2^4 = 16$ binary strings of length four, it follows that $p(E) = p(F) = \dfrac{8}{16} = \dfrac12.$
Since $$E \cap F = \{1111, 1100, 1010, 1001\},$$it follows that $p(E \cap F) = \dfrac{4}{16} = \dfrac14.$
Since $$p(E \cap F) = \dfrac14 = \dfrac12 \cdot \dfrac12 = p(E) \cdot p(F),$$it is concluded that $E$ and $F$ are independent events.

Answer a)
Since the event of selecting a red cow occurs after it is confirmed that a female cow is selected, this is a conditional probability case. Suppose $E_1$ and $F_1$ respectively represent the events of selecting a red cow and selecting a female cow.
$$\begin{aligned} p(E_1 \mid F_1) & = \dfrac{n(E_1 \cap F_1)}{n(F_1)} \\ & = \dfrac{4}{5+4+6} \\ & = \dfrac{4}{15} \end{aligned}$$Thus, the probability that the cow is red given that the selected cow is female is $\boxed{\dfrac{4}{15}}$
Answer b)
Since the event of selecting a male cow occurs after it is confirmed that a black cow is selected, this is a conditional probability case. Suppose $E_2$ and $F_2$ respectively represent the events of selecting a male cow and selecting a black cow.
$$\begin{aligned} p(E_2 \mid F_2) & = \dfrac{n(E_2 \cap F_2)}{n(F_2)} \\ & = \dfrac{5}{5+5} \\ & = \dfrac{1}{2} \end{aligned}$$Thus, the probability that the cow is red given that the selected cow is female is $\boxed{\dfrac{1}{2}}$
Answer c)
Since the event of selecting a female cow occurs after it is confirmed that a spotted cow is selected, this is a conditional probability case. Suppose $E_3$ and $F_3$ respectively represent the events of selecting a female cow and selecting a spotted cow.
First selection:
$$\begin{aligned} p(E_3 \mid F_3) & = \dfrac{n(E_3 \cap F_3)}{n(F_3)} \\ & = \dfrac{6}{2+6} \\ & = \dfrac{3}{4} \end{aligned}$$

Suppose $E_3’$ and $F_3’$ respectively represent the events of selecting a female cow and selecting a spotted cow after one spotted female cow can no longer be selected (now only $6-1=5$ cows remain).
Second selection:
$$\begin{aligned} p(E_3 \mid F_3) & = \dfrac{n(E_3 \cap F_3)}{n(F_3)} \\ & = \dfrac{5}{2+5} \\ & = \dfrac{5}{7} \end{aligned}$$Thus, the probability that the cow is red given that the selected cow is female is $\boxed{\dfrac34 \cdot \dfrac57 = \dfrac{15}{28}}$

Suppose $A, B, C,$ and $D$ respectively represent the events that components $A, B, C,$ and $D$ work.
Answer a)
Because each component operates independently, we obtain
$$\begin{aligned} p(A \cap B \cap (C \cup D)) & = p(A) \cdot p(B) \cdot p(C \cup D) \\ & = p(A) \cdot p(B) \cdot (1-p(C^c \cap D^c)) \\ & = p(A) \cdot p(B) \cdot (1-p(C^c) \cdot p(D^c)) \\ & = 0,\!9 \cdot 0,\!8 \cdot (1-(1-0,\!7) \cdot (1-0,\!8)) \\ & = 0,\!6768. \end{aligned}$$Therefore, the probability that the electrical system operates is $\boxed{0,\!6768}$
Answer b)
The event that component $C$ does not work given that the electrical system operates is a conditional event. It is known that the probability the electrical system operates is $0,\!6768.$ Thus, the probability that component $C$ does not work given that the electrical system operates is expressed as
$$\begin{aligned} p & = \dfrac{p(A \cap B \cap C^c \cap D)}{0,\!6768} \\ & = \dfrac{p(A) \cdot p(B) \cdot p(C^c) \cdot p(D)}{0,\!6768} \\ & = \dfrac{0,\!9 \cdot 0,\!8 \cdot (1-0,\!7) \cdot 0,\!8}{0,\!6768} \\ & \approx 0,\!2553. \end{aligned}$$

Suppose $A, B, C, D$ and $E$ respectively represent the events that components $A, B, C, D$ and $E$ work.
Answer a)
Because each component operates independently, we obtain
$$\begin{aligned} p((A \cap B) \cup (C \cap D \cap E)) & = 1-\left[p((A \cap B)^c \cap (C \cap D \cap E)^c)\right] \\ & = 1-\left[p((A^c \cup B^c) \cap (C^c \cup D^c \cup E^c))\right] \\ & = 1-\left[p(A^c \cup B^c) \cdot p(C^c \cup D^c \cup E^c)\right] \\ & = 1-(1-p(A \cap B))(1-p(C \cap D \cap E)) \\ & = 1-(1-p(A)p(B))(1-p(C)p(D)p(E)) \\ & = 1-(1-0,\!8 \cdot 0,\!7)(1-0,\!8 \cdot 0,\!6 \cdot 0,\!9) \\ & = 0,\!75008. \end{aligned}$$Therefore, the probability that the electrical system operates is $\boxed{0,\!75008}$
Answer b)
The event that component $A$ does not work given that the electrical system operates is a conditional event. It is known that the probability the electrical system operates is $0,\!75008.$ Thus, the probability that component $A$ does not work given that the electrical system operates is expressed as
$$\begin{aligned} p & = \dfrac{p(A^c \cap C \cap D \cap E)}{0,\!75008} \\ & = \dfrac{p(A^c) \cdot p(C) \cdot p(D) \cdot p(E)}{0,\!75008} \\ & = \dfrac{(1-0,\!8) \cdot 0,\!8 \cdot 0,\!6 \cdot 0,\!9}{0,\!75008} \\ & \approx 0,\!11519. \end{aligned}$$Answer c)
Suppose $S$ represents the event that the electrical system operates so that $p(S) = 0,\!75008.$
In this case, we will determine the value of $p(A^c \mid S^c),$ namely
$$\begin{aligned} p(A^c \mid S^c) & = \dfrac{p(A^c \cap S^c)}{p(S^c)} \\ & = \dfrac{p(A^c)(1-p(C \cap D \cap E))}{1-p(S)} \\ & = \dfrac{(0,\!2)(1-(0,\!8)(0,\!6)(0,\!9)}{1-0,\!75008} \\ & \approx 0,\!45455. \end{aligned}$$Therefore, the probability that component $A$ does not work given that the electrical system does not operate is $\boxed{0,\!45455}$