# Practice Exam Solutions for Second Midterm

#### 13 March 2007

Question text is in black, solutions in blue.

### Directions:

• Answer the problems on the exam pages.
• There are seven problems on pages 2-7, for 100 total points. Probable scale is A=93, C=63.
• If you need extra space use the back of a page.
• No books, notes, calculators, or collaboration.
• The first five questions are true/false, with five points for the correct boolean answer and up to five for a correct justification.
• Question 6 has numerical answers -- you may give your answer in the form of an expression using arithmetic operations, powers, falling powers, or the factorial function. If you give your answer using the "choose" notation, also give it using only operations on this list.

Q1: 10 points
Q2: 10 points
Q3: 10 points
Q4: 10 points
Q5: 10 points
Q6: 30 points
Q7: 20 points
Total: 100 points

• Question 1 (10): True or false with justification: Suppose I make four successive independent throws of a single six-sided die. Then the probability that I ever throw the same number twice in a row is less than or equal to 1/2.

TRUE. This event is the union of three events: (1) first and second throws are the same, (2) second and third are the same, and (3) third and fourth are the same. Each of these has probability 1/6, so their union has probability at most 3(1/6) or 1/2. The actual probability is less because of double-counting, it is 3(1/6) - 3(1/36) + 1(1/216) = 89/216, but you don't need to compute it to answer the question.

• Question 2 (10): True or false with justification: Let A, B, and C be three pairwise independent random variables over the same event space. (This means that A and B, A and C, and B and C are each independent pairs.) Then Prob(A|B) = Prob(A|C).

TRUE. Both Pr(A|B) and Pr(A|C) are equal to Pr(A) because of the independence. For example, Pr(A|B) = Pr(A∩B)/Pr(B) by definition and Pr(A∩B) = Pr(A)Pr(B) by independence, so Pr(A|B) = Pr(A). Similarly, Pr(A|C) = Pr(A).

• Question 3 (10): True or false with justification: If I collect 2n coupons, each of which is independently and uniformly chosen from one of n types, the probability that I get at least one coupon of each type is no more than 1 - 1/e2.

TRUE. The chance that I do not get type 1 is (1 - 1/n)2n, which is about equal to 1/e2 and in fact is slightly more than that. The chance of missing any of the n types is clearly still greater, so the chance of not missing any type cannot be greater than 1 - 1/e2.

• Question 4 (10): True or false with justification: It is not possible for a random variable to always take on a value that is at least one standard deviation away from its mean.

FALSE. Suppose X is 1 or -1 with equal probability. Then E(X) is 0, and E(X2) = 1 because X2 is always 1. So Var(X) = 1 - 0 = 1 and the standard deviation is the square root of the variance, also 1. Thus this X is always exactly one standard deviation from its mean.

• Question 5 (10): True or false with justification: Let X be a variable that always takes on a value less than -1, and whose mean is -2. It is not possible that Pr(X ≤ -5) = 10%.

FALSE. We might try to prove this with the Markov bound, letting Y = -1 - X so that Y is never negative. Then E(Y) = 1 and we are asked whether Pr(Y ≥ 4) could be 10%. This does not violate the Markov bound, so we expect we could construct such a variable. It's easier to first construct Y. We let Y = 4 with probability 0.1, then determine what value Y should have the rest of the time so that E(Y) = 1. If this value is z, we have that E(Y) = (0.1)(4) + (0.9)z = 1, so that (0.9)z = 0.6 and z = 2/3. Converting back, we have that X = -5/3 with probability 0.9 and X = -5 with probability 0.1.

• Question 6 (30): In college basketball, there are two different ways a player can be awarded two free throws. In a one-and-one, the player attempts her first throw and is allowed a second throw only if the first throw succeeds. In a two-shot foul, the player is allowed the second throw whether the first is successful of not. Assume that Julia's free throws are independent events, each with a 2/3 probability of success.

• (a,5) If Julia shoots a one-and-one, what is the probability that she gets two points? That she gets one point? That she gets no points? (Each successful free throw counts for one point.)

To get two points she makes the first (2/3) and the second (2/3), with probability 4/9. To get one point she makes the first (2/3) and misses the second (1/3), with probability 2/9. To get no points she misses the first (1/3) for probability 1/3. The three probabilities add to 4/9 + 2/9 + 3/9 = 1, which is good.

• (b,5) What is her expected number of points in a one-and-one?

2(4/9) + 1(2/9) + 0(1/3) = 10/9.

• (c,5) What is the variance of the number of points she gets in a one-and-one? (You may find the formula Var(X) = E(X2) - E(X)2 to be useful.)

E(X2) = 4(4/9) + 1(2/9) + 0(1/3) = 18/9 = 2. Thus Var(X) = 2 - (10/9)2 = 62/81.

• (d,5) What are the probabilities of 0, 1, or two points respectively if Julia shoots a two-shot foul?

The probability of two points is still 4/9. There are now two ways to get one point (make-miss and miss-make) with 2/9 probability each, for 4/9 total. This leaves 1/9 probability of no points (miss-miss, 1/3 times 1/3).

• (e,5) What is her expected number of points for a two-shot foul?

The expected number of points from each shot is 2/3, making 4/3 total. We could also calculate 2(4/9) + 1(4/9) + 0(1/9) = 4/3.

• (f,5) What is the variance of the number of points she gets in a two-shot foul? (Hint: This question is substantially easier than part (c) if you see how to do it.)

The two shots are now independent random variables, so their variances add. The variance of a single shot is 2/9, using the formula p(1-p) or directly, so the variance of the total points is 4/9. (We could get this by the formula np(1-p) for the variance of n Bernoulli trials with success probability p.) We could of course compute E(X2) as we did in part (c), getting 4(4/9) + 1(4/9) + 0(1/9) = 20/9, and subtract E(X)2 = 16/9 to get 4/9.

• Question 7 (20): Recall that B(n,p,i) is the probability that there are exactly i successes in a sequence of n Bernoulli trials where each trial has success probability p. Prove that for any natural number k, B(2k,0.5,k) ≥ 2-k. (Hint: Use induction on k and argue that if you get k successes in the first 2k trials, there is at least a 50% chance that you will get k+1 successes in the first 2k+2 trials.)

Base case: Let k=0, then B(0,0.5,0) = 1 because in no trials we must get no successes. The base case is true because 1 ≥ 2-0 = 1. Now assume that B(2k,0.5,k) ≥ 2-k and look at B(2k+2,0.5,k+1). One way to get k+1 successes in 2k+2 trials is to get exactly k in the first 2k trials and then exactly one of the last two. The chance of the first event is at least 2-k by the IH, and the chance of the second is 1/2 (because B(2,0.5,1) = (2 choose 1)(1/2)1(1-1/2)1). So the chance of k+1 successes is at least 2-k(1/2) = 2-(k+1), which completes the inductive case.