"Probability and Statistics" — Practice

Question 1

Two fair dice are rolled. Given that the sum is 7, what is the probability that one of the dice shows a 3?

Accepted Answer

B

Question 2

For a Poisson random variable, which statement is always true?

Accepted Answer

B

Question 3

A fair six-sided die is rolled once. What is the expected value of the number shown?

Accepted Answer

B

Question 4

Two machines produce bolts. Machine A makes 60 per cent of output with a 2 per cent defect rate; machine B makes 40 per cent with a 5 per cent defect rate. A randomly chosen bolt is defective. What is the probability it came from machine B?

Accepted Answer

Work in counts on a convenient population of 1000 bolts, which is faster and less error-prone than manipulating probabilities.

Machine A produces 600 bolts, of which 2 per cent = 12 are defective.
Machine B produces 400 bolts, of which 5 per cent = 20 are defective.

Total defectives: 12 + 20 = 32.

The probability that a defective bolt came from B is 20/32 = 5/8 = 0.625.

Notice the structure: machine B makes fewer bolts overall but contributes the majority of the defects, because its defect rate is more than double A's. The base rate and the conditional rate pull in opposite directions, and Bayes' theorem is exactly the arithmetic that resolves them.

Question 5

X and Y are independent random variables with E[X] = 3, Var(X) = 2, E[Y] = 5 and Var(Y) = 7. Find E[3X - 2Y + 4] and Var(3X - 2Y + 4).

Accepted Answer

Expectation is linear unconditionally, and the additive constant survives.

E[3X - 2Y + 4] = 3(3) - 2(5) + 4 = 9 - 10 + 4 = 3.

Variance behaves differently in two ways: the additive constant contributes nothing, and each multiplier is squared.

Var(3X - 2Y + 4) = 9 Var(X) + 4 Var(Y) = 9(2) + 4(7) = 18 + 28 = 46.

Two points are worth extracting. The coefficient minus 2 contributes plus 4, so the variances add even though the variables are subtracted. And the addition of the two variance terms is legitimate only because X and Y are independent; without that, a covariance term of minus 2 times 3 times 2 times Cov(X,Y) would appear.

Question 6

The lifetime of a component is exponentially distributed with mean 500 hours. What is the probability it survives beyond 800 hours, and beyond 1300 hours given that it has already survived 500?

Accepted Answer

The mean of an exponential is 1 over lambda, so lambda = 1/500 per hour.

The tail probability is P(X greater than t) = e to the power minus lambda t.

For 800 hours: e to the minus 800/500 = e to the minus 1.6, which is about 0.202.

For the conditional part, use memorylessness. Surviving beyond 1300 given survival past 500 requires an additional 800 hours, and the exponential has no memory of the 500 already elapsed.

So the answer is P(X greater than 800) again, about 0.202.

This is the defining property of the exponential distribution and the reason it makes queueing analysis tractable: a component that has already run for 500 hours is, statistically, exactly as good as new.

Question 7

n letters are placed at random into n addressed envelopes, one per envelope. What is the expected number of letters that land in their correct envelope?

Accepted Answer

Direct enumeration is intractable, because the events are heavily dependent: if letter 1 is correctly placed, the probabilities for the others all change.

Linearity of expectation sidesteps that entirely, because it does not require independence.

Define an indicator variable I_k which is 1 if letter k lands in its own envelope and 0 otherwise. The total number of correct placements is the sum of these n indicators.

For a single letter k, it can land in any of the n envelopes with equal probability, so P(I_k = 1) = 1/n and E[I_k] = 1/n.

By linearity, the expected total is the sum of n terms each equal to 1/n, which is exactly 1.

The answer is 1, independent of n. Whether there are 5 letters or 5 million, one correct placement is expected on average.

The instructive point is that no independence was needed at any step, and that computing the actual distribution of the number of fixed points is far harder than computing its expectation.

Question 8

Show that pairwise independence does not imply mutual independence, using two fair coin tosses.

Accepted Answer

Toss two fair coins and define three events.

A: the first coin is a head. B: the second coin is a head. C: the two coins match, that is, both heads or both tails.

Each event has probability 1/2. A and B are clearly 1/2 each, and C occurs for HH and TT out of four equally likely outcomes, so also 1/2.

Check the pairs. A and B intersect only at HH, with probability 1/4, which equals (1/2)(1/2), so they are independent. A and C intersect only at HH, again 1/4, so independent. B and C intersect only at HH, again 1/4, so independent.

All three pairs are independent.

Now check the triple. A, B and C all occur exactly at HH, so the intersection has probability 1/4. But the product of the three individual probabilities is 1/8.

Since 1/4 does not equal 1/8, the three events are not mutually independent.

The intuition is that C is completely determined by A and B together, so knowing both leaves nothing random about C, even though knowing either one alone tells you nothing.

Question 9

A random variable has mean 50 and standard deviation 5, with no distribution specified. Bound the probability that it lies outside the range 35 to 65, and compare with the normal case.

Accepted Answer

The range 35 to 65 is exactly 3 standard deviations either side of the mean, since 15 divided by 5 is 3.

Chebyshev's inequality bounds the probability of being more than k standard deviations from the mean by 1 over k squared, using only the variance and no distributional assumption.

With k = 3, the bound is 1/9, approximately 0.111.

So at most about 11 per cent of the mass lies outside the range, whatever the distribution.

If the variable were known to be normal, the true probability outside three standard deviations is about 0.3 per cent, since roughly 99.7 per cent lies within.

The gap between 11 per cent and 0.3 per cent is the price of making no assumption. Chebyshev must hold for every distribution with that variance, including highly skewed ones concentrated near the bound, so it cannot be tight for any particular well-behaved case. Its value is precisely that it applies when the stem deliberately withholds the distribution.

Practice — "Probability and Statistics"

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