Probability and Statistics

Each \(X_{i}\) is a Bernoulli variable with \(E[X_{i}] = p\) and \(\operatorname{Var}(X_{i}) = p(1 - p)\).

• By linearity of expectation, \(E[\,\widehat{p}\,] = \frac{1}{n} \cdot np = p\). So A holds and B fails; \(\frac{p}{n}\) forgets that the sum has \(n\) terms.
• The tosses are independent, so the variances add: \(\operatorname{Var}(\widehat{p}) = \frac{1}{n^{2}} \cdot np(1 - p) = \frac{p(1 - p)}{n}\). This shrinks as \(n\) grows, so C holds and D fails.

In short, \(\widehat{p}\) is an unbiased estimator of \(p\) that gets more precise with more tosses.

Answer: A, C

Drawing with replacement leaves the bag unchanged, so the draws are independent and each is black with probability \(\frac{10}{15} = \frac{2}{3}\).

Write \(S_{100} = Y_{1} + \cdots + Y_{100}\), where \(Y_{k} = 1\) if the \(k\)-th draw is black and 0 otherwise. Each \(E[Y_{k}] = \frac{2}{3}\), so by linearity of expectation

\[ E[S_{100}] = 100 \times \frac{2}{3} = \frac{200}{3} \approx 66.67 \]

This is the mean \(np\) of a Binomial\((100, \frac{2}{3})\) variable. Rounded to one decimal place it is 66.7; using the white-ball probability \(\frac{1}{3}\) by mistake would give 33.3.

Answer: 66.7

(i) A Poisson random variable with parameter \(\lambda\) has \(P(X = k) = \dfrac{e^{-\lambda}\lambda^{k}}{k!}\), and both its mean and its variance equal \(\lambda\). True.

(ii) The standard normal distribution \(N(0, 1)\) is, by definition, the normal distribution with mean 0 and variance 1; any normal variable is brought to it by \(Z = (X - \mu)/\sigma\). True.

Both statements hold.

Answer: A

\(T \cap S\) needs at least two heads and at least two tails, which takes at least four tosses. With three coins, two or more heads leave at most one tail, so the two events can never happen together.

\(T\) and \(S\) are mutually exclusive (in fact complementary, since every outcome of three tosses has a majority of heads or a majority of tails), so \(P(T \cap S) = 0\). The value 0.5 is \(P(T)\) alone.

Answer: A

The average gives the total: 50 points with mean 40 add up to \(50 \times 40 = 2000\). With the new point the total is \(2000 + 142 = 2142\) over 51 points, so the new mean is

\[\frac{2142}{51} = 42.\]

A quick check: the new point lies \(142 - 40 = 102\) above the old mean, and that excess is shared by all 51 points, raising the mean by \(102/51 = 2\). Dividing by 50 instead of 51 is the usual slip.

Answer: 42

Each throw is even with probability \(\frac{1}{2}\). Let \(E_{0}\) be the expected number of further throws at the start or just after an odd throw, and \(E_{1}\) the same just after an even throw.

• \(E_{1} = 1 + \frac{1}{2}(0) + \frac{1}{2}E_{0}\): an even throw finishes, an odd one starts over.
• \(E_{0} = 1 + \frac{1}{2}E_{1} + \frac{1}{2}E_{0}\), which gives \(E_{0} = 2 + E_{1}\).

Substituting, \(E_{0} = 3 + \frac{1}{2}E_{0}\), so \(E_{0} = 6\).

Option B (4) is the expected wait for two even throws that need not be consecutive (\(2 + 2\)); insisting that they come back to back costs two more throws on average.

Answer: C

By independence, \((X, Y)\) is uniform on the rectangle \([1, 3] \times [2, 4]\), whose area is 4, so a probability is the favourable area divided by 4.

Since \(Y\) is never below 2, \(X \ge Y\) can only happen when \(X \ge 2\). The favourable region is the triangle \(2 \le Y \le X \le 3\), with both legs of length 1:

\[ \text{area} = \int_{2}^{3} (x - 2)\,dx = \frac{1}{2}, \qquad P(X \ge Y) = \frac{1/2}{4} = \frac{1}{8} = 0.125 \]

Answer: 0.125

For an exponential random variable with rate \(\lambda\):

\[ E(X) = \int_{0}^{\infty} x\lambda e^{-\lambda x}\,dx = \frac{1}{\lambda}, \qquad Var(X) = E(X^{2}) - E(X)^{2} = \frac{2}{\lambda^{2}} - \frac{1}{\lambda^{2}} = \frac{1}{\lambda^{2}} \]

The condition \(5E(X) = Var(X)\) becomes \(\frac{5}{\lambda} = \frac{1}{\lambda^{2}}\). Multiplying by \(\lambda^{2}\), which is allowed because \(\lambda > 0\), gives \(5\lambda = 1\), so \(\lambda = 0.2\).

Taking the mean as \(\lambda\) instead of \(1/\lambda\) is the usual slip; it would give \(\lambda = 5\).

Answer: 0.2

First, \(P(T) = 1 - P(\overline{T}) = 0.4\). By the law of total probability,

\[ P(S) = P(S|T)P(T) + P(S|\overline{T})P(\overline{T}) = 0.3 \times 0.4 + 0.6 \times 0.6 = 0.12 + 0.36 = 0.48 \]

Bayes' theorem then reverses the conditioning:

\[ P(T|S) = \frac{P(S|T)P(T)}{P(S)} = \frac{0.12}{0.48} = 0.25 \]

Note that \(P(T|S) = 0.25\) differs from the given \(P(S|T) = 0.3\); confusing the two is the common trap.

Answer: 0.25

The marks went to all because the printed function is not a valid density: over its region it integrates to \(\int_{0}^{2}\int_{0}^{x} 2xy\,dy\,dx = \int_{0}^{2} x^{3}\,dx = 4\), not 1.

The conditional expectation can still be found as printed, because a constant factor cancels in a conditional density. For a fixed \(x\),

\[ f_{Y|X}(y|x) = \frac{2xy}{\int_{0}^{x} 2xy\,dy} = \frac{2xy}{x^{3}} = \frac{2y}{x^{2}}, \quad 0 < y < x \]

\[ E[Y|X = x] = \int_{0}^{x} y \cdot \frac{2y}{x^{2}}\,dy = \frac{2x}{3}, \qquad E[Y|X = 1.5] = \frac{2 \times 1.5}{3} = 1 \]

Answer: 1