Probability Theory Questions and Answers

Q: Suppose X is an absolutely continuous random variable with probability density function \(f(.)\) and characteristic function \(\phi(.)\). Suppose \(\phi(.)\) is a Riemann integrable on R. Then :

The inversion formula is \(f(x) = \frac{1}{2\pi}\int_{R} e^{-iux}\phi(u)\,du\).With \(\phi(u) = E[e^{iuX}] = \int e^{iux}f(x)\,dx\), the Fourier inversion theorem says that when \(\phi\) is integrable, X has a bounded continuous density given by\[ \begin{aligned} &f(x) \\ &= \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-iux}\phi(u)\,du \end{aligned} \]The exponent must have the opposite sign to the one in \(\phi\), which rules out (A); the factor \(1/2\pi\) is needed (B lacks it); and the integral runs over the whole line, not just \((0, \infty)\) as in (D).

Q: Suppose P(\(x\), \(y\)) denotes the joint probability mass function of two random variables X and Y where :P(0, 0) = 0.4, P(0, 1) = 0.2, P(1, 0) = 0.1, P(1, 1) = 0.3.What is the conditional probability of X = 0, given that Y = 1 ?

\(\mathrm{P}(X = 0 \mid Y = 1) = \) 2/5.\[ \begin{aligned} P(Y = 1) &= P(0, 1) + P(1, 1) = 0.2 + 0.3 = 0.5 \\ P(X = 0 \mid Y = 1) &= \frac{P(0, 1)}{P(Y = 1)} = \frac{0.2}{0.5} = \frac{2}{5} \end{aligned} \]1/5 is the unconditional probability 0.2 taken without dividing by \(P(Y = 1)\), and 3/5 is \(P(X = 1 \mid Y = 1)\). The probability cannot be 0, because the pair (0, 1) has positive probability.

Q: If 3 balls are randomly drawn from a bowl containing 6 white and 5 black balls, what is the probability that one of the balls is white and other two black ?

The probability is 4/11.There are \(\binom{11}{3} = 165\) equally likely ways to draw 3 balls. One white and two black can be chosen in \(\binom{6}{1}\binom{5}{2} = 6 \times 10 = 60\) ways:\[ P = \frac{60}{165} = \frac{4}{11} \]As a check, the other outcomes are three white (20), two white and one black (75) and three black (10); with 60 these add up to 165. 8/11 and 3/11 do not correspond to any count here, and 2/11 = 30/165 would need only 30 favourable ways.

Q: In a class of 100 students, 40 students got 60 marks and 60 students got 40 marks out of 100. Hence the mean, median and mode of the class marks are respectively.

The mean, median and mode are 48, 40 and 40.\[ \begin{aligned} \text{Mean} &= \frac{40 \times 60 + 60 \times 40}{100} \\ &= \frac{4800}{100} = 48 \end{aligned} \]Arranged in order, the 60 marks of 40 come first, so the 50th and 51st values, and hence the median, are 40. The mode is the most frequent value, 40, which 60 students obtained.24 halves the mean by dividing by 200 instead of 100, and 60 is the less frequent mark, so it cannot be the mode or the median.

Q: The coefficient of variation of a random variable X is 1.2. If \(Y = \dfrac{X}{4}\), the coefficient of variation of Y is :

The coefficient of variation of Y is 1.2.The coefficient of variation is \(\sigma/\mu\). Dividing X by 4 divides both the standard deviation and the mean by 4:\[ \begin{aligned} CV_{Y} &= \frac{\sigma_{X}/4}{\mu_{X}/4} \\ &= \frac{\sigma_{X}}{\mu_{X}} = 1.2 \end{aligned} \]The CV is unchanged by any change of scale, which is why it is used to compare variability across different units. 0.3 divides the CV by 4, 0.6 by 2, and 2.4 doubles it; each forgets that the mean is scaled too.

Q: Suppose P(A/B) = 0.4 and P(B) = 0.8. Which of the following is a possible value for P(A) ?

A possible value is 0.5.\(P(A \cap B) = P(A \mid B)P(B) = 0.4 \times 0.8 = 0.32\). Then\[ \begin{aligned} 0.32 = P(A \cap B) &\le P(A) \\ &\le P(A \cap B) + P(B^{c}) \\ &= 0.32 + 0.2 = 0.52 \end{aligned} \]so \(P(A)\) must lie in \([0.32, 0.52]\). Only 0.5 is in that range. 0.3 and 0.1 are smaller than \(P(A \cap B)\), and 0.6 would need more of A outside B than \(P(B^{c}) = 0.2\) allows.

Q: Let X be a random variable with \(E[|X|^{3}] = 3\).Then :

The conclusion is \(P[|X| \le 2] \ge 5/8\).By Markov's inequality applied to \(|X|^{3}\):\[ \begin{aligned} P[|X| \gt 2] &= P[|X|^{3} \gt 8] \\ &\le \frac{E|X|^{3}}{8} = \frac{3}{8} \end{aligned} \]so \(P[|X| \le 2] \ge 1 - 3/8 = 5/8\).The bound goes the other way from options A and B, which put upper limits on the probability, and the moment information is enough to say something, so C is wrong.

Q: Let the joint distribution of (X, Y) be given by :Y \ X012301/101/201/201/2013/201/53/205/20Then \(P[X \le 1]\) is :

\(P[X \le 1] = \) 1/2.Add the probabilities in the columns X = 0 and X = 1 over both values of Y:\[ \begin{aligned} &\frac{1}{10} + \frac{3}{20} + \frac{1}{20} + \frac{1}{5} \\ &= \frac{2 + 3 + 1 + 4}{20} = \frac{10}{20} = \frac{1}{2} \end{aligned} \]As a check, all eight entries add to 1. 3/20 and 1/20 are single cells of the table, and 1/5 is the cell P(X = 1, Y = 1) alone.

Q: Which of the following is a characteristic function of a random variable ?

The characteristic function is \(\Phi(t) = \cos(t/2)\).If X takes the values \(\pm\frac{1}{2}\) with probability \(\frac{1}{2}\) each, then \(E[e^{itX}] = \frac{1}{2}(e^{it/2} + e^{-it/2}) = \cos(t/2)\).The others fail. A characteristic function satisfies \(\Phi(0) = 1\) and \(|\Phi(t)| \le 1\): \(t/(1 + t^{2})\) gives \(\Phi(0) = 0\), and \(2/(1 + 2t)\) gives 2 at \(t = 0\). \(\cos(t^{2})\) passes those tests but has \(\Phi''(0) = 0\), which would mean \(E[X^{2}] = 0\), so X = 0 and \(\Phi \equiv 1\), a contradiction.

Q: Suppose E, F and G are 3 events such that \(P(E \cap F \cap G) = P(E).P(F).P(G)\) and \(P(E \cap F \cap G) = P(E).P(F \cap G)\).Consider the statements :(I) F and G are independent events(II) E and \(F^{C} \cup G^{C}\) are independent events(III) \(E^{C}\) and \(F \cap G\) are independent events(IV) E, F and G are mutually independent events

The correct statements are I, II and III.The two equations give \(P(E)P(F \cap G) = P(E)P(F)P(G)\). Assuming \(P(E) \gt 0\), this yields \(P(F \cap G) = P(F)P(G)\), so F and G are independent (I).The second equation says E and \(F \cap G\) are independent. Independence is preserved under complements, so \(E^{c}\) and \(F \cap G\) are independent (III), and E and \((F \cap G)^{c} = F^{c} \cup G^{c}\) are independent (II).IV does not follow: mutual independence also needs \(P(E \cap F) = P(E)P(F)\) and \(P(E \cap G) = P(E)P(G)\), which are not given. So options including IV, or omitting I, are wrong.

51.

Suppose X is an absolutely continuous random variable with probability density function \(f(.)\) and characteristic function \(\phi(.)\). Suppose \(\phi(.)\) is a Riemann integrable on R. Then :

A.\(f(x) = \frac{1}{2\pi}\int_{R} e^{iux}\, \phi(u)du\)
B.\(f(x) = \int_{R} e^{-iux}\, \phi(u)du\)
C.\(f(x) = \frac{1}{2\pi}\int_{R} e^{-iux}\, \phi(u)du\)
D.\(f(x) = \frac{1}{2\pi}\int_{0}^{\infty} e^{-iux}\, \phi(u)du\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

The inversion formula is \(f(x) = \frac{1}{2\pi}\int_{R} e^{-iux}\phi(u)\,du\).

With \(\phi(u) = E[e^{iuX}] = \int e^{iux}f(x)\,dx\), the Fourier inversion theorem says that when \(\phi\) is integrable, X has a bounded continuous density given by

\[ \begin{aligned} &f(x) \\ &= \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-iux}\phi(u)\,du \end{aligned} \]

The exponent must have the opposite sign to the one in \(\phi\), which rules out (A); the factor \(1/2\pi\) is needed (B lacks it); and the integral runs over the whole line, not just \((0, \infty)\) as in (D).

Discuss in Forum

52.

Suppose P(\(x\), \(y\)) denotes the joint probability mass function of two random variables X and Y where :

P(0, 0) = 0.4, P(0, 1) = 0.2, P(1, 0) = 0.1, P(1, 1) = 0.3.

What is the conditional probability of X = 0, given that Y = 1 ?

A.\(\dfrac{1}{5}\)
B.\(\dfrac{3}{5}\)
C.\(\dfrac{2}{5}\)
D.0

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

\(\mathrm{P}(X = 0 \mid Y = 1) = \) 2/5.

\[ \begin{aligned} P(Y = 1) &= P(0, 1) + P(1, 1) = 0.2 + 0.3 = 0.5 \\ P(X = 0 \mid Y = 1) &= \frac{P(0, 1)}{P(Y = 1)} = \frac{0.2}{0.5} = \frac{2}{5} \end{aligned} \]

1/5 is the unconditional probability 0.2 taken without dividing by \(P(Y = 1)\), and 3/5 is \(P(X = 1 \mid Y = 1)\). The probability cannot be 0, because the pair (0, 1) has positive probability.

Discuss in Forum

53.

If 3 balls are randomly drawn from a bowl containing 6 white and 5 black balls, what is the probability that one of the balls is white and other two black ?

A.\(\dfrac{3}{11}\)
B.\(\dfrac{4}{11}\)
C.\(\dfrac{8}{11}\)
D.\(\dfrac{2}{11}\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

The probability is 4/11.

There are \(\binom{11}{3} = 165\) equally likely ways to draw 3 balls. One white and two black can be chosen in \(\binom{6}{1}\binom{5}{2} = 6 \times 10 = 60\) ways:

\[ P = \frac{60}{165} = \frac{4}{11} \]

As a check, the other outcomes are three white (20), two white and one black (75) and three black (10); with 60 these add up to 165. 8/11 and 3/11 do not correspond to any count here, and 2/11 = 30/165 would need only 30 favourable ways.

Discuss in Forum

54.

In a class of 100 students, 40 students got 60 marks and 60 students got 40 marks out of 100. Hence the mean, median and mode of the class marks are respectively.

A.48, 40 and 60
B.48, 40 and 40
C.24, 60 and 40
D.24, 40 and 60

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

The mean, median and mode are 48, 40 and 40.

\[ \begin{aligned} \text{Mean} &= \frac{40 \times 60 + 60 \times 40}{100} \\ &= \frac{4800}{100} = 48 \end{aligned} \]

Arranged in order, the 60 marks of 40 come first, so the 50th and 51st values, and hence the median, are 40. The mode is the most frequent value, 40, which 60 students obtained.

24 halves the mean by dividing by 200 instead of 100, and 60 is the less frequent mark, so it cannot be the mode or the median.

Discuss in Forum

55.

The coefficient of variation of a random variable X is 1.2. If \(Y = \dfrac{X}{4}\), the coefficient of variation of Y is :

A.0.3
B.0.6
C.2.4
D.1.2

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

The coefficient of variation of Y is 1.2.

The coefficient of variation is \(\sigma/\mu\). Dividing X by 4 divides both the standard deviation and the mean by 4:

\[ \begin{aligned} CV_{Y} &= \frac{\sigma_{X}/4}{\mu_{X}/4} \\ &= \frac{\sigma_{X}}{\mu_{X}} = 1.2 \end{aligned} \]

The CV is unchanged by any change of scale, which is why it is used to compare variability across different units. 0.3 divides the CV by 4, 0.6 by 2, and 2.4 doubles it; each forgets that the mean is scaled too.

Discuss in Forum

56.

Suppose P(A/B) = 0.4 and P(B) = 0.8. Which of the following is a possible value for P(A) ?

A.0.3
B.0.5
C.0.6
D.0.1

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

A possible value is 0.5.

\(P(A \cap B) = P(A \mid B)P(B) = 0.4 \times 0.8 = 0.32\). Then

\[ \begin{aligned} 0.32 = P(A \cap B) &\le P(A) \\ &\le P(A \cap B) + P(B^{c}) \\ &= 0.32 + 0.2 = 0.52 \end{aligned} \]

so \(P(A)\) must lie in \([0.32, 0.52]\). Only 0.5 is in that range. 0.3 and 0.1 are smaller than \(P(A \cap B)\), and 0.6 would need more of A outside B than \(P(B^{c}) = 0.2\) allows.

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57.

Let X be a random variable with \(E[|X|^{3}] = 3\).

Then :

A.\(P[|X| \le 2] \le \dfrac{3}{8}\)
B.\(P[|X| \le 2] \le \dfrac{3}{5}\)
C.One cannot say anything about \(P[|X| \le 2]\) from the given information
D.\(P[|X| \le 2] \ge \dfrac{5}{8}\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

The conclusion is \(P[|X| \le 2] \ge 5/8\).

By Markov's inequality applied to \(|X|^{3}\):

\[ \begin{aligned} P[|X| \gt 2] &= P[|X|^{3} \gt 8] \\ &\le \frac{E|X|^{3}}{8} = \frac{3}{8} \end{aligned} \]

so \(P[|X| \le 2] \ge 1 - 3/8 = 5/8\).

The bound goes the other way from options A and B, which put upper limits on the probability, and the moment information is enough to say something, so C is wrong.

Discuss in Forum

58.

Let the joint distribution of (X, Y) be given by :

Y \ X	0	1	2	3
0	1/10	1/20	1/20	1/20
1	3/20	1/5	3/20	5/20

Then \(P[X \le 1]\) is :

A.3/20
B.1/20
C.1/2
D.1/5

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

\(P[X \le 1] = \) 1/2.

Add the probabilities in the columns X = 0 and X = 1 over both values of Y:

\[ \begin{aligned} &\frac{1}{10} + \frac{3}{20} + \frac{1}{20} + \frac{1}{5} \\ &= \frac{2 + 3 + 1 + 4}{20} = \frac{10}{20} = \frac{1}{2} \end{aligned} \]

As a check, all eight entries add to 1. 3/20 and 1/20 are single cells of the table, and 1/5 is the cell P(X = 1, Y = 1) alone.

Discuss in Forum

59.

Which of the following is a characteristic function of a random variable ?

A.\(\Phi(t) = \cos(t^{2})\)
B.\(\Phi(t) = \cos(t/2)\)
C.\(\Phi(t) = \dfrac{2}{1 + 2t}\)
D.\(\Phi(t) = \dfrac{t}{1 + t^{2}}\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

The characteristic function is \(\Phi(t) = \cos(t/2)\).

If X takes the values \(\pm\frac{1}{2}\) with probability \(\frac{1}{2}\) each, then \(E[e^{itX}] = \frac{1}{2}(e^{it/2} + e^{-it/2}) = \cos(t/2)\).

The others fail. A characteristic function satisfies \(\Phi(0) = 1\) and \(|\Phi(t)| \le 1\): \(t/(1 + t^{2})\) gives \(\Phi(0) = 0\), and \(2/(1 + 2t)\) gives 2 at \(t = 0\). \(\cos(t^{2})\) passes those tests but has \(\Phi''(0) = 0\), which would mean \(E[X^{2}] = 0\), so X = 0 and \(\Phi \equiv 1\), a contradiction.

Discuss in Forum

60.

Suppose E, F and G are 3 events such that \(P(E \cap F \cap G) = P(E).P(F).P(G)\) and \(P(E \cap F \cap G) = P(E).P(F \cap G)\).

Consider the statements :

(I) F and G are independent events
(II) E and \(F^{C} \cup G^{C}\) are independent events
(III) \(E^{C}\) and \(F \cap G\) are independent events
(IV) E, F and G are mutually independent events

A.Only statements I, II and III are correct
B.Only statements II, III and IV are correct
C.Only statements II and III are correct
D.Only statement I is correct

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

The correct statements are I, II and III.

The two equations give \(P(E)P(F \cap G) = P(E)P(F)P(G)\). Assuming \(P(E) \gt 0\), this yields \(P(F \cap G) = P(F)P(G)\), so F and G are independent (I).

The second equation says E and \(F \cap G\) are independent. Independence is preserved under complements, so \(E^{c}\) and \(F \cap G\) are independent (III), and E and \((F \cap G)^{c} = F^{c} \cup G^{c}\) are independent (II).

IV does not follow: mutual independence also needs \(P(E \cap F) = P(E)P(F)\) and \(P(E \cap G) = P(E)P(G)\), which are not given. So options including IV, or omitting I, are wrong.

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