Probability Theory Questions and Answers

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.

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.

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.

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.

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.

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.

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.

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.

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.

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 \(P(R)\).