Probability and Statistics

\(E[X|Y]\) is itself a random variable: a function of \(Y\) that gives the average of \(X\) for each value \(Y\) can take. Averaging it over the distribution of \(Y\) recovers the overall average of \(X\). For discrete variables,

\(E[E[X|Y]] = \sum_{y} E[X|Y = y]\,P(Y = y) = \sum_{y}\sum_{x} x\,P(X = x, Y = y) = \sum_{x} x\,P(X = x) = E[X]\).

This is the law of total expectation (the tower property). Option A is still random, not a fixed number, and B and D fail in general: if \(Y\) is independent of \(X\), then \(E[X|Y] = E[X]\), so the result is \(E[X]\) whatever \(E[Y]\) is.

Answer: C

The median \(m\) is the point where the CDF reaches one half, \(F_{X}(m) = \frac{1}{2}\). Since 3 lies in the middle piece (\(t \leq 3 \leq 4\)),

\[\frac{3 - t}{4 - t} = \frac{1}{2} \;\Rightarrow\; 6 - 2t = 4 - t \;\Rightarrow\; t = 2.\]

Check: with \(t = 2\), \(X\) is uniform on \([2, 4]\), whose midpoint is 3. The other options fail; for example, \(t = 0\) gives \(F_{X}(3) = \frac{3}{4}\), so the median would be 2 instead.

Answer: A

Use \(E[Z] = 0\) and \(E[Z^{2}] = \operatorname{Var}(Z) = 1\).

• \(E[X] = aE[Z] + b = b\), so \(b = 1\).
• \(X - E[X] = aZ\), so \(E[(X - E[X])Z] = aE[Z^{2}] = a\), which gives \(a = -2\).
• Check: \(E[(X - E[X])^{2}] = a^{2}E[Z^{2}] = 4\) holds. This condition alone cannot fix the sign of \(a\); the second one does.

Option C has the right \(a\) but the wrong mean, while B and D fail the condition \(E[(X - E[X])Z] = -2\).

Answer: A

An exponential random variable with mean \(\frac{1}{\lambda}\) has rate \(\lambda\) and tail probability \(P(X \geq x) = e^{-\lambda x}\) for \(x \geq 0\). So

\[e^{-2\lambda} = 0.25 = \frac{1}{4} \;\Rightarrow\; 2\lambda = \ln 4 = 2\ln 2 \;\Rightarrow\; \lambda = \ln 2.\]

Check: \(e^{-2\ln 2} = 2^{-2} = 0.25\). Option B is the value of \(2\lambda\), not \(\lambda\), and option D is negative (\(\ln 0.25 = -\ln 4\)), which is impossible for a rate.

Answer: A

Let \(W\) be the event that the ball is white. The chance of white from each box is \(\frac{1}{3}\), \(\frac{2}{3}\) and \(\frac{3}{6} = \frac{1}{2}\). By the law of total probability,

\[P(W) = \frac{1}{2}\cdot\frac{1}{3} + \frac{1}{6}\cdot\frac{2}{3} + \frac{1}{3}\cdot\frac{1}{2} = \frac{3}{18} + \frac{2}{18} + \frac{3}{18} = \frac{4}{9}.\]

By Bayes' theorem,

\[P(\text{Box-2} \mid W) = \frac{\frac{1}{6}\cdot\frac{2}{3}}{\frac{4}{9}} = \frac{1/9}{4/9} = \frac{1}{4} = 0.25.\]

Box-2 has the highest share of white balls, but it is the least likely box to be chosen, which keeps its posterior low.

Answer: 0.25

Standardising a normal variable gives \(Z \sim N(0, 1)\), so \(E[Z^{2}] = \operatorname{Var}(Z) = 1\). The fourth moment of a standard normal is \(E[Z^{4}] = 3\): its moment generating function \(e^{t^{2}/2}\) has \(t^{4}\) coefficient \(\frac{1}{8}\), and \(4! \cdot \frac{1}{8} = 3\). Therefore

\[\operatorname{Var}(Y) = E[Z^{4}] - \left(E[Z^{2}]\right)^{2} = 3 - 1 = 2.\]

Equivalently, \(Y = Z^{2}\) is chi-squared with 1 degree of freedom, whose variance is twice the degrees of freedom. Option C is \(E[Z^{4}]\) alone, not the variance.

Answer: B

\(X^{2} \le 0.25\) means \(-0.5 \le X \le 0.5\). The CDF has no jumps (it rises continuously from 0 at \(x = -1\) to 1 at \(x = 1\)), so

\[P(-0.5 \le X \le 0.5) = F_X(0.5) - F_X(-0.5) = \frac{(1.5)^{2} - (0.5)^{2}}{4} = \frac{2}{4} = 0.5.\]

Two slips lead to the distractors: \(F_X(0.5) = 0.5625\) alone forgets the lower limit (D), and reading the event as \(|X| \le 0.25\) gives \(F_X(0.25) - F_X(-0.25) = 0.25\) (B).

Answer: C

\(Y\) is a sum of 300 independent Bernoulli variables with \(\theta = 0.25\), so

• \(E[Y] = 300 \times 0.25 = 75\);
• \(\operatorname{Var}(Y) = 300 \times 0.25 \times 0.75 = 56.25\), a standard deviation of \(7.5\).

By the Central Limit Theorem, \(\frac{Y - 75}{7.5}\) is approximately standard normal. The limits standardise to \(\frac{60 - 75}{7.5} = -2\) and \(\frac{90 - 75}{7.5} = 2\), so

\[P(60 \le Y \le 90) \approx \phi(2) - \phi(-2).\]

Option D puts the raw values of \(Y\) into \(\phi\) without standardising.

Answer: A

A mean of \(\frac{1}{\ln 10}\) means the rate is \(\lambda = \ln 10\), so \(P(X \ge t) = e^{-t \ln 10} = 10^{-t}\) for \(t \ge 0\).

\(Y = \ell\) happens exactly when \(\ell \le X \lt \ell + 1\), so

\[P(Y = \ell) = 10^{-\ell} - 10^{-(\ell + 1)} = (0.1)^{\ell}\,(1 - 0.1).\]

This has the form \(q^{\ell}(1 - q)\) with \(q = 0.1\), so the floor of an exponential variable is geometric. Option D, \(0.434 \approx \frac{1}{\ln 10}\), is the mean of \(X\), not \(q\).

Answer: A

Use the complement: the event fails only when no die shows a 1.

Each die avoids 1 with probability \(\frac{5}{6}\), and the 100 throws are independent, so \(P(\text{no 1 at all}) = \left(\frac{5}{6}\right)^{100}\). Therefore

\[P(A) = 1 - \left(\frac{5}{6}\right)^{100}.\]

Option D is the probability of the complement. Options A and B fail because both \(A\) and its complement can happen, so \(P(A)\) lies strictly between 0 and 1, though very close to 1.

Answer: C