Probability Theory Questions and Answers

Q: The joint probability mass function PX,Y of the random variables X and Y is as given in the following table :PX,Y(x, y)Y = 1Y = 2Y = 3Y = 4PX(x)X = 10.020.040.060.080.20X = 20.030.060.090.120.30X = 30.050.100.150.200.50PY(y)0.100.200.300.40Then, which of the following statements is true ?

X and Y are uncorrelated.Every cell of the table equals the product of its marginals: \(0.02 = 0.20 \times 0.10\), \(0.06 = 0.30 \times 0.20\), \(0.15 = 0.50 \times 0.30\), and so on for all twelve cells. So X and Y are independent, and independent variables have zero covariance, hence are uncorrelated (and not positively correlated).The means are \(E(X) = 0.2 + 0.6 + 1.5 = 2.3\) and \(E(Y) = 0.1 + 0.4 + 0.9 + 1.6 = 3.0\), so \(E(Y) \lt E(X)\) is false. And \(P(X \gt 2, Y \gt 2) = 0.15 + 0.20 = 0.35\), while \(1 - P(X \le 2, Y \le 3) = 1 - 0.30 = 0.70\), so the stated identity fails.

Q: In order to detect an error, a computer program is being tested independently by three different testers. The probabilities with which the testers can detect the error are 0.2, 0.4 and 0.5 respectively. Then, which of the following statements is false ?

The false statement is All the above are false, because the first three are true.The miss probabilities are 0.8, 0.6 and 0.5.\[ \begin{aligned} P(\text{none}) &= 0.8 \times 0.6 \times 0.5 = 0.24 \\ P(\text{exactly one}) &= 0.06 + 0.16 + 0.24 = 0.46 \end{aligned} \]where the three terms are \(0.2 \times 0.6 \times 0.5\), \(0.8 \times 0.4 \times 0.5\) and \(0.8 \times 0.6 \times 0.5\). Then\[ P(\text{at least two}) = 1 - 0.24 - 0.46 = 0.30 \]So statements A, B and C are all true, and the statement claiming they are all false is itself the false one.

Q: If E and F are events with \(P(E) \lt P(F)\) and \(P(E \cap F) \gt 0\), then :

None of the implications holds.The conditions \(P(E) \lt P(F)\) and \(P(E \cap F) \gt 0\) say nothing about one event containing the other. Roll a die with E = {1, 2} and F = {2, 3, 4}. Then \(P(E) = \frac{1}{3} \lt \frac{1}{2} = P(F)\) and \(E \cap F = \{2\}\) has positive probability.If 1 is rolled, E occurs but F does not, so E does not imply F. If 3 is rolled, F occurs but E does not, so F does not imply E. If 3 is rolled, E fails but F occurs, so non-occurrence of E does not imply non-occurrence of F.

Q: Suppose X is an arbitrary random variable. Which of the following statements is true ?

The true statement is P[|X| ≥ 3] ≤ E(X4)/81.This is Markov's inequality applied to the non-negative variable \(X^{4}\): \(|X| \ge 3\) is the same event as \(X^{4} \ge 81\), so\[ P(|X| \ge 3) = P(X^{4} \ge 81) \le \frac{E(X^{4})}{81} \]It bounds the tail from above, so the reversed inequality is wrong. The statements about \(P(|X| \le 3)\) are not what Markov's inequality gives; for example, if \(X = 0\) always, \(P(|X| \le 3) = 1\) while \(E(X^{4})/81 = 0\), so the first option fails.

Q: Suppose \(\phi(t) = \dfrac{1}{(1 + t^{2})}\) is a characteristic function of a random variable X. Then E(X) and Var(X) are respectively given by :

The mean and variance are 0 and 2.\(\phi(t) = \frac{1}{1 + t^{2}}\) is the characteristic function of the standard Laplace distribution with density \(\frac{1}{2}e^{-|x|}\). Directly from the characteristic function, using the expansion near 0:\[ \phi(t) = 1 - t^{2} + t^{4} - \ldots \]Comparing with \(\phi(t) = 1 + itE(X) - \frac{t^{2}}{2}E(X^{2}) + \ldots\), there is no t term, so \(E(X) = 0\), and \(\frac{1}{2}E(X^{2}) = 1\), so \(E(X^{2}) = 2\) and \(Var(X) = 2\).The variance 1 is the coefficient of \(t^{2}\) taken without the factor \(\frac{1}{2}\), and the options with mean \(\frac{1}{2}\) ignore that \(\phi\) is real and even, which forces a symmetric distribution.

Q: Suppose the probability space \((\Omega, A, P)\) is defined as follows : \(\Omega = [0, 1]\), A is a sigma field of subsets of \(\Omega\) and P is a Lebesgue measure. Suppose a sequence \(\{X_{n}, n \ge 1\}\) of random variables is defined as \(X_{n}(w) = e^{w} + w^{n}\), \(n \ge 1\) and a random variable X is defined as \(X(w) = e^{w}\) on (\(\Omega\), A, P). Which of the following is not true as \(n \to \infty\) ?

The statement that is not true is Xn → X pointwise.For \(0 \le w \lt 1\), \(w^{n} \to 0\), so \(X_{n}(w) \to e^{w} = X(w)\). At \(w = 1\), \(X_{n}(1) = e + 1\) for every n, which does not tend to \(X(1) = e\). So convergence fails at one point and is not pointwise on all of \(\Omega\).The single point {1} has Lebesgue measure 0, so \(X_{n} \to X\) almost surely, and almost sure convergence implies convergence in probability and in distribution. Those three statements are true.

Q: The upper quartile of the data is 7.75 and coefficient of quartile deviation 0.55. The interquartile range is :

The interquartile range is 5.50.The coefficient of quartile deviation is \(\frac{Q_{3} - Q_{1}}{Q_{3} + Q_{1}}\). With \(Q_{3} = 7.75\):\[ \begin{aligned} 7.75 - Q_{1} &= 0.55(7.75 + Q_{1}) \\ 0.45 \times 7.75 &= 1.55\, Q_{1} \\ Q_{1} &= 2.25 \end{aligned} \]So the interquartile range is \(Q_{3} - Q_{1} = 7.75 - 2.25 = 5.50\), and the quartile deviation (half of it) is 2.75.Check: \(\frac{5.5}{10} = 0.55\). The other values do not satisfy this ratio.

Q: The population of Nicosia is 75% Greek and 25% Turkish. 20% of the Greeks and 10% of the Turks speak English. A visitor to the town meets someone who speaks English. What is the probability that he is a Greek ?

The probability is \(\frac{6}{7}\).By Bayes' theorem, compare the two routes to meeting an English speaker:\[ \begin{aligned} P(G \cap E) &= 0.75 \times 0.20 = 0.150 \\ P(T \cap E) &= 0.25 \times 0.10 = 0.025 \\ P(G \mid E) &= \frac{0.150}{0.175} = \frac{6}{7} \end{aligned} \]\(\frac{1}{7}\) is the probability that the English speaker is Turkish. \(\frac{3}{7}\), \(\frac{4}{7}\) and \(\frac{5}{7}\) come from mixing up the weights, for example ignoring the population shares.

Q: Let \(F(x) = P[X \le x]\) be the distribution function of a random variable X.Define \(D_{n} = \left\{u \mid F(u) - F(u-) \ge \frac{1}{n}\right\}\), \(n = 1, 2, 3, \ldots\)Then :

The true statement is that \(D_{n}\) cannot contain more than n elements.\(F(u) - F(u-)\) is the jump of F at u, which equals \(P(X = u)\). Every point of \(D_{n}\) carries probability at least \(\frac{1}{n}\), and the total probability is 1, so there can be at most n such points.The other statements fail. For a continuous distribution (normal, say) F has no jumps, so every \(D_{n}\) is empty. \(D_{2}\) has 2 elements only when there are two atoms of mass at least \(\frac{1}{2}\) each, as in a fair coin, and not 'always'. \(D_{1}\) is not always empty: for a degenerate distribution, the single atom has mass 1.

Q: A man is known to speak truth 3 out of 4 times. He throws a fair die and reports it is a six. Then, the probability that it is actually six is equal to :

The probability that it is actually a six is 3/8.Let S be 'a six was thrown' and R 'he reports a six'. He reports six when it is a six and he tells the truth, or when it is not and he lies:\[ \begin{aligned} P(S \cap R) &= \tfrac{1}{6} \times \tfrac{3}{4} = \tfrac{3}{24} \\ P(S^{c} \cap R) &= \tfrac{5}{6} \times \tfrac{1}{4} = \tfrac{5}{24} \\ P(S \mid R) &= \frac{3/24}{8/24} = \frac{3}{8} \end{aligned} \]\(\frac{5}{8}\) is the probability that he lied, and \(\frac{1}{8}\) is the unconditional \(P(S \cap R)\) without dividing by \(P(R)\).

81.

The joint probability mass function P_X,Y of the random variables X and Y is as given in the following table :

P_X,Y(x, y)	Y = 1	Y = 2	Y = 3	Y = 4	P_X(x)
X = 1	0.02	0.04	0.06	0.08	0.20
X = 2	0.03	0.06	0.09	0.12	0.30
X = 3	0.05	0.10	0.15	0.20	0.50
P_Y(y)	0.10	0.20	0.30	0.40

Then, which of the following statements is true ?

A.\(E(Y) \lt E(X)\)
B.\(P(X \gt 2, Y \gt 2) = 1 - P(X \le 2, Y \le 3)\)
C.X and Y are uncorrelated
D.X and Y are positively correlated

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

X and Y are uncorrelated.

Every cell of the table equals the product of its marginals: \(0.02 = 0.20 \times 0.10\), \(0.06 = 0.30 \times 0.20\), \(0.15 = 0.50 \times 0.30\), and so on for all twelve cells. So X and Y are independent, and independent variables have zero covariance, hence are uncorrelated (and not positively correlated).

The means are \(E(X) = 0.2 + 0.6 + 1.5 = 2.3\) and \(E(Y) = 0.1 + 0.4 + 0.9 + 1.6 = 3.0\), so \(E(Y) \lt E(X)\) is false. And \(P(X \gt 2, Y \gt 2) = 0.15 + 0.20 = 0.35\), while \(1 - P(X \le 2, Y \le 3) = 1 - 0.30 = 0.70\), so the stated identity fails.

Discuss in Forum

82.

In order to detect an error, a computer program is being tested independently by three different testers. The probabilities with which the testers can detect the error are 0.2, 0.4 and 0.5 respectively. Then, which of the following statements is false ?

A.The probability that the error will not be detected is 0.24
B.The probability that exactly one tester will find the error is 0.46
C.The probability that at least two testers will find the error is 0.30
D.All the above are false

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: D

The false statement is All the above are false, because the first three are true.

The miss probabilities are 0.8, 0.6 and 0.5.

\[ \begin{aligned} P(\text{none}) &= 0.8 \times 0.6 \times 0.5 = 0.24 \\ P(\text{exactly one}) &= 0.06 + 0.16 + 0.24 = 0.46 \end{aligned} \]

where the three terms are \(0.2 \times 0.6 \times 0.5\), \(0.8 \times 0.4 \times 0.5\) and \(0.8 \times 0.6 \times 0.5\). Then

\[ P(\text{at least two}) = 1 - 0.24 - 0.46 = 0.30 \]

So statements A, B and C are all true, and the statement claiming they are all false is itself the false one.

Discuss in Forum

83.

If E and F are events with \(P(E) \lt P(F)\) and \(P(E \cap F) \gt 0\), then :

A.Occurrence of E \(\Rightarrow\) occurrence of F
B.Occurrence of F \(\Rightarrow\) occurrence of E
C.Non-occurrence of E \(\Rightarrow\) non-occurrence of F
D.None of the above implications hold

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: D

None of the implications holds.

The conditions \(P(E) \lt P(F)\) and \(P(E \cap F) \gt 0\) say nothing about one event containing the other. Roll a die with E = {1, 2} and F = {2, 3, 4}. Then \(P(E) = \frac{1}{3} \lt \frac{1}{2} = P(F)\) and \(E \cap F = \{2\}\) has positive probability.

If 1 is rolled, E occurs but F does not, so E does not imply F. If 3 is rolled, F occurs but E does not, so F does not imply E. If 3 is rolled, E fails but F occurs, so non-occurrence of E does not imply non-occurrence of F.

Discuss in Forum

84.

Suppose X is an arbitrary random variable. Which of the following statements is true ?

A.\(P[|X| \le 3] \le \dfrac{E(X^{4})}{81}\)
B.\(P[|X| \le 3] \ge \dfrac{E(X^{4})}{81}\)
C.\(P[|X| \ge 3] \le \dfrac{E(X^{4})}{81}\)
D.\(P[|X| \ge 3] \ge \dfrac{E(X^{4})}{81}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The true statement is P[|X| ≥ 3] ≤ E(X⁴)/81.

This is Markov's inequality applied to the non-negative variable \(X^{4}\): \(|X| \ge 3\) is the same event as \(X^{4} \ge 81\), so

\[ P(|X| \ge 3) = P(X^{4} \ge 81) \le \frac{E(X^{4})}{81} \]

It bounds the tail from above, so the reversed inequality is wrong. The statements about \(P(|X| \le 3)\) are not what Markov's inequality gives; for example, if \(X = 0\) always, \(P(|X| \le 3) = 1\) while \(E(X^{4})/81 = 0\), so the first option fails.

Discuss in Forum

85.

Suppose \(\phi(t) = \dfrac{1}{(1 + t^{2})}\) is a characteristic function of a random variable X. Then E(X) and Var(X) are respectively given by :

A.\(\dfrac{1}{2}, \dfrac{1}{8}\)
B.0, 2
C.0, 1
D.\(\dfrac{1}{2}, \dfrac{1}{16}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

The mean and variance are 0 and 2.

\(\phi(t) = \frac{1}{1 + t^{2}}\) is the characteristic function of the standard Laplace distribution with density \(\frac{1}{2}e^{-|x|}\). Directly from the characteristic function, using the expansion near 0:

\[ \phi(t) = 1 - t^{2} + t^{4} - \ldots \]

Comparing with \(\phi(t) = 1 + itE(X) - \frac{t^{2}}{2}E(X^{2}) + \ldots\), there is no t term, so \(E(X) = 0\), and \(\frac{1}{2}E(X^{2}) = 1\), so \(E(X^{2}) = 2\) and \(Var(X) = 2\).

The variance 1 is the coefficient of \(t^{2}\) taken without the factor \(\frac{1}{2}\), and the options with mean \(\frac{1}{2}\) ignore that \(\phi\) is real and even, which forces a symmetric distribution.

Discuss in Forum

86.

Suppose the probability space \((\Omega, A, P)\) is defined as follows : \(\Omega = [0, 1]\), A is a sigma field of subsets of \(\Omega\) and P is a Lebesgue measure. Suppose a sequence \(\{X_{n}, n \ge 1\}\) of random variables is defined as \(X_{n}(w) = e^{w} + w^{n}\), \(n \ge 1\) and a random variable X is defined as \(X(w) = e^{w}\) on (\(\Omega\), A, P). Which of the following is not true as \(n \to \infty\) ?

A.\(X_{n} \to X\) almost surely
B.\(X_{n} \to X\) in probability
C.\(X_{n} \to X\) in distribution
D.\(X_{n} \to X\) pointwise

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: D

The statement that is not true is X_n → X pointwise.

For \(0 \le w \lt 1\), \(w^{n} \to 0\), so \(X_{n}(w) \to e^{w} = X(w)\). At \(w = 1\), \(X_{n}(1) = e + 1\) for every n, which does not tend to \(X(1) = e\). So convergence fails at one point and is not pointwise on all of \(\Omega\).

The single point {1} has Lebesgue measure 0, so \(X_{n} \to X\) almost surely, and almost sure convergence implies convergence in probability and in distribution. Those three statements are true.

Discuss in Forum

87.

The upper quartile of the data is 7.75 and coefficient of quartile deviation 0.55. The interquartile range is :

A.5.00
B.5.25
C.5.50
D.5.75

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: C

The interquartile range is 5.50.

The coefficient of quartile deviation is \(\frac{Q_{3} - Q_{1}}{Q_{3} + Q_{1}}\). With \(Q_{3} = 7.75\):

\[ \begin{aligned} 7.75 - Q_{1} &= 0.55(7.75 + Q_{1}) \\ 0.45 \times 7.75 &= 1.55\, Q_{1} \\ Q_{1} &= 2.25 \end{aligned} \]

So the interquartile range is \(Q_{3} - Q_{1} = 7.75 - 2.25 = 5.50\), and the quartile deviation (half of it) is 2.75.

Check: \(\frac{5.5}{10} = 0.55\). The other values do not satisfy this ratio.

Discuss in Forum

88.

The population of Nicosia is 75% Greek and 25% Turkish. 20% of the Greeks and 10% of the Turks speak English. A visitor to the town meets someone who speaks English. What is the probability that he is a Greek ?

A.\(\frac{3}{7}\)
B.\(\frac{4}{7}\)
C.\(\frac{5}{7}\)
D.\(\frac{6}{7}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: D

The probability is \(\frac{6}{7}\).

By Bayes' theorem, compare the two routes to meeting an English speaker:

\[ \begin{aligned} P(G \cap E) &= 0.75 \times 0.20 = 0.150 \\ P(T \cap E) &= 0.25 \times 0.10 = 0.025 \\ P(G \mid E) &= \frac{0.150}{0.175} = \frac{6}{7} \end{aligned} \]

\(\frac{1}{7}\) is the probability that the English speaker is Turkish. \(\frac{3}{7}\), \(\frac{4}{7}\) and \(\frac{5}{7}\) come from mixing up the weights, for example ignoring the population shares.

Discuss in Forum

89.

Let \(F(x) = P[X \le x]\) be the distribution function of a random variable X.

Define \(D_{n} = \left\{u \mid F(u) - F(u-) \ge \frac{1}{n}\right\}\), \(n = 1, 2, 3, \ldots\)

Then :

A.\(D_{n}\) cannot contain more than n elements
B.\(D_{n}\) is not empty set for all \(n \ge 1\)
C.\(D_{2}\) always contains 2 elements
D.\(D_{1}\) is always empty set

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The true statement is that \(D_{n}\) cannot contain more than n elements.

\(F(u) - F(u-)\) is the jump of F at u, which equals \(P(X = u)\). Every point of \(D_{n}\) carries probability at least \(\frac{1}{n}\), and the total probability is 1, so there can be at most n such points.

The other statements fail. For a continuous distribution (normal, say) F has no jumps, so every \(D_{n}\) is empty. \(D_{2}\) has 2 elements only when there are two atoms of mass at least \(\frac{1}{2}\) each, as in a fair coin, and not 'always'. \(D_{1}\) is not always empty: for a degenerate distribution, the single atom has mass 1.

Discuss in Forum

90.

A man is known to speak truth 3 out of 4 times. He throws a fair die and reports it is a six. Then, the probability that it is actually six is equal to :

A.1/8
B.5/8
C.2/7
D.3/8

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: D

The probability that it is actually a six is 3/8.

Let S be 'a six was thrown' and R 'he reports a six'. He reports six when it is a six and he tells the truth, or when it is not and he lies:

\[ \begin{aligned} P(S \cap R) &= \tfrac{1}{6} \times \tfrac{3}{4} = \tfrac{3}{24} \\ P(S^{c} \cap R) &= \tfrac{5}{6} \times \tfrac{1}{4} = \tfrac{5}{24} \\ P(S \mid R) &= \frac{3/24}{8/24} = \frac{3}{8} \end{aligned} \]

\(\frac{5}{8}\) is the probability that he lied, and \(\frac{1}{8}\) is the unconditional \(P(S \cap R)\) without dividing by \(P(R)\).

Probability Theory Questions and Answers

Topics

Other Subjects

Tags: Probability Theory Questions and Answers

About Us

Probability Theory Questions and Answers

Topics

Other Subjects

Tags: Probability Theory Questions and Answers

About Us

Stay informed about all important Dates & Notifications