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Stat 110 Midterm
Prof. Joe Blitzstein
October 12, 2011
This exam is closed book and closed notes, except for two standard-sized sheets of paper (8.5” by 11”) which can have notes on both sides. No copying, cheating, collaboration, calculators, computers, or cell phones are allowed. Show your work. Answers should be exact unless an approximation is asked for. All parts will be weighted equally within each problem. The last page contains a table of important distributions, and the page before that can be used for scratch work or for extra space. If you want any work done on the extra page or on backs of pages to be graded, mention where to look in big letters with a box around them, on the page with the question. Good luck (an appropriate expression for this course)!
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- A group of 360 people are going to be split into 120 teams of 3 (where the order of teams and the order within a team don’t matter).
(a) How many ways are there to do this (simplified, in terms of factorials)?
(b) The 360 people consist of 180 married couples. A random split into teams of 3 is chosen, with all possible splits equally likely. Find the expected number of teams containing married couples. (You can leave your answer in terms of binomial coe cients and a product of a few terms, but you should not have summations or complicated expressions in your final answer.)
- Let X ⇠ Bin(100, 0 .9). For each of the following parts, construct an example showing that it is possible, or explain clearly why it is impossible. In this problem, Y is a random variable on the same probability space as X; note that X and Y are not necessarily independent.
(a) Is it possible to have Y ⇠ Pois(0.01) with P (X Y ) = 1?
(b) Is it possible to have a Y ⇠ Bin(100, 0 .5) with P (X Y ) = 1?
(c) Is it possible to have Y ⇠ Bin(100, 0 .5) with P (X Y ) = 1?
Extra page 1. This page can be used for scratch work or as extra space. If you want work in the extra space (or on backs of pages) to be graded, indicate this very clearly on the page with the corresponding problem.
Extra page 2. This page can be used for scratch work or as extra space. If you want work in the extra space (or on backs of pages) to be graded, indicate this very clearly on the page with the corresponding problem.
Stat 110 Midterm Solutions, Fall 2011
Prof. Joe Blitzstein
- A new treatment for a disease is being tested, to see whether it is better than the standard treatment. The existing treatment is e↵ective on 50% of patients. It is believed initially that there is a 2/3 chance that the new treatment is e↵ective on 60% of patients, and a 1/3 chance that the new treat- ment is e↵ective on 50% of patients. In a pilot study done with 20 randomly selected patients, the new treatment is e↵ective for 15 of the patients.
(a) Given this information, what is the probability that the new treatment is better than the standard treatment (as an unsimplified number)?
Let B be the event that the new treatment is better than the standard treatment and let X be the number of people in the study for whom the new treatment is e↵ective. By Bayes’ Rule and the Law of Total Probability,
P (B|X = 15) =
P (X = 15|B)P (B)
P (X = 15|B)P (B) + P (X = 15|B c^ )P (B c^ )
=
15
15
15
(This problem is a form of the “coin with a random bias” problem.)
(b) A second study is then done, with a set of 20 new random patients. Given the results of the first study, what is the conditional PMF for how many of the new patients the new treatment is e↵ective on? (Don’t simplify; it may help to let p be the answer to (a), and give your answer in terms of p.)
Let Y be how many of the new patients the new treatment is e↵ective for nd p = P (B|X = 15) be the answer from (a). Then for k 2 { 0 , 1 ,... , 20 },
P (Y = k|X = 15) = P (Y = k|X = 15, B)P (B|X = 15) + P (Y = k|X = 15, B c^ )P (B c^ |X = 15)
= P (Y = k|B)P (B|X = 15) + P (Y = k|B c^ )P (B c^ |X = 15)
=
k
(0.6) k^ (0.4) 20 k^ p +
k
(0.5) 20 (1 p).
(This distribution is not Binomial. As in the coin with a random bias prob- lem, the individual outcomes are conditionally independent but not inde- pendent. Given the true probability of e↵ectiveness of the new treatment, the pilot study is irrelevant and the distribution is Binomial, but without knowing that, we have a mixture of two Binomials.)
- A group of 360 people are going to be split into 120 teams of 3 (where the order of teams and the order within a team don’t matter).
(a) How many ways are there to do this (simplified, in terms of factorials)?
Similarly to the Strategic Practice problem about partnerships, imagine lin- ing the people up and saying the first 3 are a team, the next 3 are a team, etc. This overcounts by a factor of (3!)^120 · 120! since the order within teams and the order of teams don’t matter. So the number of ways is
360! 6 120 · 120!
(b) The 360 people consist of 180 married couples, and a random split into teams is chosen, with all possible splits equally likely. Find the expected number of teams containing married couples. (You can leave your answer in terms of binomial coe cients and a product of a few terms, but you should not have summations or complicated expressions in your final answer.)
Let I (^) j be the indicator for the jth team having a married couple (taking the teams to be chosen one at a time, or with respect to a random ordering). By symmetry and linearity, the desired quantity is 120E(I 1 ). We have
E(I 1 ) = P (first team has a married couple) =
3
since the first team is equally likely to be any 3 of the people, and to have a married couple on the team we need to choose a couple and then any third person. So the expected value is
120 · 180 · 358 (^360) 3
(This simplifies to (^360120) · 359 ·^180 · 358 ·^358 / 6 = 360359. Another way to find the probability that the first team has a married couple is to note that any particular pair in the team has probability 3591 of being married to each other, so since there are 3 disjoint possibilities the probability is 3593 .)
- Athletes compete one at a time at the high jump. Let X (^) j be how high the jth jumper jumped, with X 1 , X 2 ,... i.i.d. with ln(X (^) j ) ⇠ N (μ, 2 ). We say that the jth jumper is “best in recent memory” if he or she jumps higher than the previous 2 jumpers (for j 3; the first 2 jumpers don’t qualify).
(a) Find the expected number of best in recent memory jumpers among the 3rd through nth jumpers (simplify).
Let I (^) j be the indicator of the jth jumper being best in recent memory, for each j 3. By symmetry, E(I (^) j ) = 1/3 (similarly to examples from class and the homework). By linearity, the desired expected value is (n 2)/3.
(b) Let Aj be the event that the jth jumper is the best in recent memory. Find P (A 3 \ A 4 ), P (A 3 ), and P (A 4 ) (simplify). Are A 3 and A 4 independent?
The event A 3 \ A 4 occurs if and only if the ranks of the first 4 jumps are 4 , 3 , 2 , 1 or 3, 4 , 2 , 1 (where 1 denotes the best of the first 4 jumps, etc.). Since all orderings are equally likely,
P (A 3 \ A 4 ) =
(Alternatively, note that A 3 \A 4 occurs if and only if the 3rd and 4th jumpers set records, and apply the result of that homework problem.) As in (a), we have P (A 3 ) = P (A 4 ) = 1/ 3. So P (A 3 \ A 4 ) 6 = P (A 3 )P (A 4 ), which shows that A 3 and A 4 are not independent.
(c) Find the variance of X (^) j (you can leave your answer in terms of integrals).
Let Y (^) j = ln(X (^) j ), so that X (^) j = e Y^ j^. By LOTUS,
Var(X (^) j ) = E(e 2 Y^ j^ ) (E(e Y^ j^ )) 2
=
p 2 ⇡
Z 1
e 2 y^ e (y μ)^
(^2) /(2 2 ) dy
p 2 ⇡
Z 1
e y^ e (y μ)^
(^2) /(2 2 ) dy
