Probability Theory Questions and Answers

Q: Suppose (X, Y) have joint pdf \(f(x, y)\), where :\[ f(x, y) = \begin{cases} 2 & 0 \lt x \lt y \lt 1 \\ 0 & \text{otherwise} \end{cases} \]Then \(E(X/y)\) is given by :

\(E(X|Y = y) =\) \(y/2\).The marginal of Y is \(f_{Y}(y) = \int_{0}^{y} 2\,dx = 2y\) for \(0 \lt y \lt 1\). The conditional density of X given \(Y = y\) is \(2/(2y) = 1/y\) on \((0, y)\), the uniform distribution on \((0, y)\). Its mean is the midpoint:\[ E(X|Y = y) = \int_{0}^{y} \frac{x}{y}\,dx = \frac{y}{2} \]\(2y\) is the marginal density of Y, not a conditional mean. \(y\) would be the upper end of the range, and 1 would put X at the top of the triangle, outside \(x \lt y\).

Q: Let X be a degenerate random variable, at \(X = c\). Then characteristic function of X at 't' is :

The characteristic function is \(\exp(itc)\).The characteristic function is \(\varphi(t) = E(e^{itX})\). A degenerate variable takes the single value \(c\) with probability 1, so\[ \varphi(t) = e^{itc} \]Its modulus is 1 for every \(t\), and \(\varphi(0) = 1\) as for any characteristic function, which rules out the constants \(c\) and 0. \(\exp(-itc)\) is the complex conjugate, the characteristic function of \(-X\), or of X under the opposite sign convention, which is not the standard definition.

Q: Two unbiased dice are thrown. The probability that both dice show the same number is :

The probability is \(\frac{1}{6}\).Two dice give \(6 \times 6 = 36\) equally likely outcomes. Both show the same number in exactly 6 of them: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6). So\[ P = \frac{6}{36} = \frac{1}{6} \]Equivalently, whatever the first die shows, the second matches it with probability 1/6. The value 1/36 is the probability of one particular double, such as two sixes. The value 11/36 is the chance that at least one die shows a given number, such as at least one six, and 1/3 corresponds to no natural count here.

Q: Suppose A and B are mutually exclusive events having probabilities P(A) = 0.25, P(B) = 0.35. What is the probability that A occurs but B does not ?

The probability is 0.25.Mutually exclusive means \(A \cap B = \emptyset\), so whenever A occurs B does not. Hence\[ P(A \cap B^{c}) = P(A) - P(A \cap B) = 0.25 - 0 = 0.25 \]0.1 is \(P(B) - P(A)\), 0.6 is \(P(A \cup B)\), and 0.4 is \(1 - P(A \cup B)\), the chance that neither occurs. None of these is the event 'A but not B'.

Q: Two boys and two girls are lined up randomly in a row. What is the probability that the girls and boys alternate ?

The probability is 1/3.Treat the four children as distinct. There are \(4! = 24\) orders. An alternating row is BGBG or GBGB, and each pattern can be filled in \(2! \times 2! = 4\) ways, so there are 8 alternating orders:\[ P = \frac{8}{24} = \frac{1}{3} \]Counting patterns only gives the same: of the \(\binom{4}{2} = 6\) equally likely boy-girl patterns, 2 alternate. 1/2 and 2/3 come from counting just one pattern or its complement wrongly.

Q: For two random variables \(x\) and \(y\) covariance between \(2x\) and \(y\), \(\mathrm{cov}(2x, y) = 4\). Hence, \(\mathrm{cov}(5x - 2, 2y - 5)\) is :

\(\mathrm{cov}(5x - 2, 2y - 5) =\) 20.Covariance ignores added constants and is bilinear in scale. From \(\mathrm{cov}(2x, y) = 2\,\mathrm{cov}(x, y) = 4\), \(\mathrm{cov}(x, y) = 2\). Then\[ \begin{aligned} &\mathrm{cov}(5x - 2, 2y - 5) \\ &= 5 \cdot 2 \cdot \mathrm{cov}(x, y) = 10 \cdot 2 = 20 \end{aligned} \]40 multiplies the given value 4 by 10 without first dividing by 2, 10 forgets that \(\mathrm{cov}(x, y)\) is 2 rather than 1, and 0 would wrongly treat the shifts as removing the relationship.

Q: If \(P(A | B) = P(B)\), then :

The answer is \(P^{2}(B) \le P(A)\).\(P(A | B) = P(A \cap B)/P(B) = P(B)\) gives \(P(A \cap B) = P^{2}(B)\). Since \(A \cap B \subseteq A\),\[ P^{2}(B) = P(A \cap B) \le P(A) \]The others need not hold. Take \(P(A) = 0.7\), \(P(B) = 0.4\) and \(P(A \cap B) = 0.16\), which is consistent and gives \(P(A | B) = 0.4 = P(B)\). Then \(P(B | A) = 0.16/0.7 \approx 0.23 \ne P(A)\); A and B are not independent, since \(P(A)P(B) = 0.28 \ne 0.16\); and \(P^{2}(A) = 0.49 \gt P(B)\).

Q: The probability mass function of a random variable X is given by :\(k\)0123\(P[X = k]\)0.20.1\(p\)\(q\)where \(p \ge 0\) and \(q \ge 0\).Which of the following is feasible ?

The feasible value is E[X] = 2.1.The probabilities must sum to 1, so \(p + q = 1 - 0.3 = 0.7\). Then\[ \begin{aligned} E[X] &= 0.1 + 2p + 3q \\ &= 0.1 + 2(0.7) + q = 1.5 + q \end{aligned} \]With \(0 \le q \le 0.7\), \(E[X]\) ranges over [1.5, 2.2]. 2.1 is reached with \(q = 0.6\), \(p = 0.1\). 1.2 is below the minimum, and 2.8 and 3.4 exceed the maximum (3.4 is even above the largest value X can take).

Q: Let \(X_{1}, X_{2}, X_{3}\) be independent identically distributed random variables with probability mass function given below :\(k\)\(-1\)012\(P[X = k]\)0.20.30.10.4Then \(P[X_{1} + X_{2} = 0/X_{2} = 1, X_{3} = 2]\) is :

The probability is 0.2.Given \(X_{2} = 1\), the event \(X_{1} + X_{2} = 0\) is the event \(X_{1} = -1\). The variables are independent, so conditioning on \(X_{2}\) and \(X_{3}\) does not change the distribution of \(X_{1}\):\[ \begin{aligned} &P[X_{1} = -1 \mid X_{2} = 1, X_{3} = 2] \\ &= P[X_{1} = -1] = 0.2 \end{aligned} \]The condition on \(X_{3}\) is irrelevant. 0.3 is \(P[X = 0]\), the value for \(X_{1} + X_{2} = 0\) only if \(X_{2}\) were 0, and 0.11 and 0.13 come from multiplying probabilities that conditioning removes.

Q: Let \(M_{V}(t)\) denote the mgf of a random variable V.Suppose X and Y are two independent r.v.s. whose mgf exists. Let \(Z = X + Y\) and \(W = XY\). Then which of the following is true for all \(t \in (-h, h)\) for some \(h \gt 0\).

The answer is \(M_{Z}(t) = M_{X}(t)M_{Y}(t)\) and \(M_{W}(t) = \int M_{Y}(tx)f_{X}(x)\,dx\).For the sum, independence gives \(E[e^{t(X + Y)}] = E[e^{tX}]E[e^{tY}]\). For the product, condition on X:\[ \begin{aligned} M_{W}(t) &= E[e^{tXY}] = E\big[E[e^{(tX)Y} \mid X]\big] \\ &= E[M_{Y}(tX)] \end{aligned} \]which is \(\int M_{Y}(tx)f_{X}(x)\,dx\) when X has a density. (Strictly, this needs \(M_{Y}(tx)\) to be finite for the relevant \(x\).)The mgf of a product is not the sum or the product of the mgfs, and the mgf of a sum is the product, not the sum, so B, C and D each contain a false half.

21.

Suppose (X, Y) have joint pdf \(f(x, y)\), where :

\[ f(x, y) = \begin{cases} 2 & 0 \lt x \lt y \lt 1 \\ 0 & \text{otherwise} \end{cases} \]

Then \(E(X/y)\) is given by :

A.\(2y\)
B.\(y/2\)
C.1
D.\(y\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

\(E(X|Y = y) =\) \(y/2\).

The marginal of Y is \(f_{Y}(y) = \int_{0}^{y} 2\,dx = 2y\) for \(0 \lt y \lt 1\). The conditional density of X given \(Y = y\) is \(2/(2y) = 1/y\) on \((0, y)\), the uniform distribution on \((0, y)\). Its mean is the midpoint:

\[ E(X|Y = y) = \int_{0}^{y} \frac{x}{y}\,dx = \frac{y}{2} \]

\(2y\) is the marginal density of Y, not a conditional mean. \(y\) would be the upper end of the range, and 1 would put X at the top of the triangle, outside \(x \lt y\).

Discuss in Forum

22.

Let X be a degenerate random variable, at \(X = c\). Then characteristic function of X at 't' is :

A.\(c\)
B.0
C.\(\exp(itc)\)
D.\(\exp(-itc)\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The characteristic function is \(\exp(itc)\).

The characteristic function is \(\varphi(t) = E(e^{itX})\). A degenerate variable takes the single value \(c\) with probability 1, so

\[ \varphi(t) = e^{itc} \]

Its modulus is 1 for every \(t\), and \(\varphi(0) = 1\) as for any characteristic function, which rules out the constants \(c\) and 0. \(\exp(-itc)\) is the complex conjugate, the characteristic function of \(-X\), or of X under the opposite sign convention, which is not the standard definition.

Discuss in Forum

23.

Two unbiased dice are thrown. The probability that both dice show the same number is :

A.\(\frac{1}{3}\)
B.\(\frac{11}{36}\)
C.\(\frac{1}{6}\)
D.\(\frac{1}{36}\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The probability is \(\frac{1}{6}\).

Two dice give \(6 \times 6 = 36\) equally likely outcomes. Both show the same number in exactly 6 of them: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6). So

\[ P = \frac{6}{36} = \frac{1}{6} \]

Equivalently, whatever the first die shows, the second matches it with probability 1/6. The value 1/36 is the probability of one particular double, such as two sixes. The value 11/36 is the chance that at least one die shows a given number, such as at least one six, and 1/3 corresponds to no natural count here.

Discuss in Forum

24.

Suppose A and B are mutually exclusive events having probabilities P(A) = 0.25, P(B) = 0.35. What is the probability that A occurs but B does not ?

A.0.1
B.0.25
C.0.4
D.0.6

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

The probability is 0.25.

Mutually exclusive means \(A \cap B = \emptyset\), so whenever A occurs B does not. Hence

\[ P(A \cap B^{c}) = P(A) - P(A \cap B) = 0.25 - 0 = 0.25 \]

0.1 is \(P(B) - P(A)\), 0.6 is \(P(A \cup B)\), and 0.4 is \(1 - P(A \cup B)\), the chance that neither occurs. None of these is the event 'A but not B'.

Discuss in Forum

25.

Two boys and two girls are lined up randomly in a row. What is the probability that the girls and boys alternate ?

A.2/3
B.1/2
C.1/3
D.3/4

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

The probability is 1/3.

Treat the four children as distinct. There are \(4! = 24\) orders. An alternating row is BGBG or GBGB, and each pattern can be filled in \(2! \times 2! = 4\) ways, so there are 8 alternating orders:

\[ P = \frac{8}{24} = \frac{1}{3} \]

Counting patterns only gives the same: of the \(\binom{4}{2} = 6\) equally likely boy-girl patterns, 2 alternate. 1/2 and 2/3 come from counting just one pattern or its complement wrongly.

Discuss in Forum

26.

For two random variables \(x\) and \(y\) covariance between \(2x\) and \(y\), \(\mathrm{cov}(2x, y) = 4\). Hence, \(\mathrm{cov}(5x - 2, 2y - 5)\) is :

A.40
B.0
C.10
D.20

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

\(\mathrm{cov}(5x - 2, 2y - 5) =\) 20.

Covariance ignores added constants and is bilinear in scale. From \(\mathrm{cov}(2x, y) = 2\,\mathrm{cov}(x, y) = 4\), \(\mathrm{cov}(x, y) = 2\). Then

\[ \begin{aligned} &\mathrm{cov}(5x - 2, 2y - 5) \\ &= 5 \cdot 2 \cdot \mathrm{cov}(x, y) = 10 \cdot 2 = 20 \end{aligned} \]

40 multiplies the given value 4 by 10 without first dividing by 2, 10 forgets that \(\mathrm{cov}(x, y)\) is 2 rather than 1, and 0 would wrongly treat the shifts as removing the relationship.

Discuss in Forum

27.

If \(P(A | B) = P(B)\), then :

A.\(P(B | A) = P(A)\)
B.A and B are independent events
C.\(P^{2}(A) \le P(B)\)
D.\(P^{2}(B) \le P(A)\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

The answer is \(P^{2}(B) \le P(A)\).

\(P(A | B) = P(A \cap B)/P(B) = P(B)\) gives \(P(A \cap B) = P^{2}(B)\). Since \(A \cap B \subseteq A\),

\[ P^{2}(B) = P(A \cap B) \le P(A) \]

The others need not hold. Take \(P(A) = 0.7\), \(P(B) = 0.4\) and \(P(A \cap B) = 0.16\), which is consistent and gives \(P(A | B) = 0.4 = P(B)\). Then \(P(B | A) = 0.16/0.7 \approx 0.23 \ne P(A)\); A and B are not independent, since \(P(A)P(B) = 0.28 \ne 0.16\); and \(P^{2}(A) = 0.49 \gt P(B)\).

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28.

The probability mass function of a random variable X is given by :

\(k\)	0	1	2	3
\(P[X = k]\)	0.2	0.1	\(p\)	\(q\)

where \(p \ge 0\) and \(q \ge 0\).

Which of the following is feasible ?

A.E[X] = 1.2
B.E[X] = 2.1
C.E[X] = 2.8
D.E[X] = 3.4

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

The feasible value is E[X] = 2.1.

The probabilities must sum to 1, so \(p + q = 1 - 0.3 = 0.7\). Then

\[ \begin{aligned} E[X] &= 0.1 + 2p + 3q \\ &= 0.1 + 2(0.7) + q = 1.5 + q \end{aligned} \]

With \(0 \le q \le 0.7\), \(E[X]\) ranges over [1.5, 2.2]. 2.1 is reached with \(q = 0.6\), \(p = 0.1\). 1.2 is below the minimum, and 2.8 and 3.4 exceed the maximum (3.4 is even above the largest value X can take).

Discuss in Forum

29.

Let \(X_{1}, X_{2}, X_{3}\) be independent identically distributed random variables with probability mass function given below :

\(k\)	\(-1\)	0	1	2
\(P[X = k]\)	0.2	0.3	0.1	0.4

Then \(P[X_{1} + X_{2} = 0/X_{2} = 1, X_{3} = 2]\) is :

A.0.11
B.0.13
C.0.2
D.0.3

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

The probability is 0.2.

Given \(X_{2} = 1\), the event \(X_{1} + X_{2} = 0\) is the event \(X_{1} = -1\). The variables are independent, so conditioning on \(X_{2}\) and \(X_{3}\) does not change the distribution of \(X_{1}\):

\[ \begin{aligned} &P[X_{1} = -1 \mid X_{2} = 1, X_{3} = 2] \\ &= P[X_{1} = -1] = 0.2 \end{aligned} \]

The condition on \(X_{3}\) is irrelevant. 0.3 is \(P[X = 0]\), the value for \(X_{1} + X_{2} = 0\) only if \(X_{2}\) were 0, and 0.11 and 0.13 come from multiplying probabilities that conditioning removes.

Discuss in Forum

30.

Let \(M_{V}(t)\) denote the mgf of a random variable V.

Suppose X and Y are two independent r.v.s. whose mgf exists. Let \(Z = X + Y\) and \(W = XY\). Then which of the following is true for all \(t \in (-h, h)\) for some \(h \gt 0\).

A.\(M_{Z}(t) = M_{X}(t)\,M_{Y}(t)\) and \(M_{W}(t) = \int M_{Y}(tx)\,f_{X}(x)\,dx\)
B.\(M_{Z}(t) = M_{X}(t)\,M_{Y}(t)\) and \(M_{W}(t) = M_{X}(t) + M_{Y}(t)\)
C.\(M_{Z}(t) = M_{X}(t) + M_{Y}(t)\) and \(M_{W}(t) = M_{X}(t)\,M_{Y}(t)\)
D.\(M_{Z}(t) = \int M_{Y}(t + x)\,f_{X}(x)\,dx\) and \(M_{W}(t) = M_{X}(t)\,M_{Y}(t)\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: A

The answer is \(M_{Z}(t) = M_{X}(t)M_{Y}(t)\) and \(M_{W}(t) = \int M_{Y}(tx)f_{X}(x)\,dx\).

For the sum, independence gives \(E[e^{t(X + Y)}] = E[e^{tX}]E[e^{tY}]\). For the product, condition on X:

\[ \begin{aligned} M_{W}(t) &= E[e^{tXY}] = E\big[E[e^{(tX)Y} \mid X]\big] \\ &= E[M_{Y}(tX)] \end{aligned} \]

which is \(\int M_{Y}(tx)f_{X}(x)\,dx\) when X has a density. (Strictly, this needs \(M_{Y}(tx)\) to be finite for the relevant \(x\).)

The mgf of a product is not the sum or the product of the mgfs, and the mgf of a sum is the product, not the sum, so B, C and D each contain a false half.

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