STAT 515 - Statistics I Midterm Exam One Name: Instruction: You can use a calculator that has no connection to the Internet. Books, notes, cellphones, and computers are NOT allowed in the test. There are seven problems in total. Please show all your work. Scratch papers are provided on Pages 9-11. A formula sheet is provided in the last page. You may tear them out for convenience.. The exam is 1.5 hours. Good Luck! Problem Points 1 /15 2 /15 3 /15 4 /15 5 /15 6 /15 7 /10 Total /100
2 1. (15 pts) A fair coin is tossed three times, and each time the upper face is observed. Denote head by H and tail by T. (a) (3 pts) List all the sample points for this experiment. (b) (4 pts) Let A denote the event that exactly two heads are observed, and B the event that at least one head is observed. List the sample points in both A and B. (c) (8 pts) Find (i) P (A) = (ii) P (B) = (iii) P (A B) = (iv) P (Ā B) =
2. (15 pts, 3 pts each) If A and B are two independent events such that P (A) = 0.5 and P (B) = 0.2, find the following probabilities: (a) P (A B) = 3 (b) P (A B) = (c) P (Ā B) = (d) P (A A B) = (e) P (A B) =
4 3. (15 pts) Five cards are dealt from a standard 52-card playing deck. Note that there are four suits of cards: Hearts, Spades, Club and Diamond, and each suit consists of 13 cards from A to K. Find the following probabilities. (You can leave the formula in terms of combinatorial numbers, without computing the exact numbers). (a) (5 pts) What is the probability that we draw two hearts, one spade and two diamonds? (b) (5 pts) What is the probability that we draw a flush? ( Flush means that all five cards are of the same suite.) (c) (5 pts) What is the probability that we draw a straight flush? ( Straight flush means that all five cards are of the same suite and in sequence, for instance, 8 9 10 J Q. Also, note that Aces can play either high or low, that is, 10 J Q K A and A 2 3 4 5 are both straight flush.)
4. (15 pts) A population of voters contains 70% of Republicans and 30% of Democrats. It is reported that 50% of the Republicans and 80% of the Democrats favor an election issue. A person is chosen at random from this population. (a) (8 pts) What is the probability that this person does NOT favor the issue. 5 (b) (7 pts) If the chosen person is found to favor the issue, what is the conditional probability that this person is a Democrat?
6 5. (15 pts) A supervisor in a manufacturing plant has four men and six women working for him. He wants to choose two workers randomly for a special job. Let Y denote the number of women in his selection. (a) (9 pts) Find the probability distribution function p(y) for Y. (b) (3 pts) Find the expected value E(Y ). (c) (3 pts) Find the variance V (Y ).
6. (15 pts) The odds are two-to-one that, when Alice and Bob play tennis, Alice wins. Suppose that Alice and Bob play five matches, and the result of each match would not affect the others. (a) (4 pts) Let Y be the number of matches that Alice wins. Clearly Y follows a binomial distribution b(n, p), find the parameters n and p. 7 (b) (7 pts) What is the probability that Alice wins at least one match? (c) (4 pts) What is the expected number of matches that Alice wins?
8 7. (10 pts) Suppose that only 30% of the applicants for a job opening are qualified. Applicants are interviewed sequentially and are selected at random. (a) (5 pts) Find the expected number of interviews needed to meet the first qualified applicant. (b) (5 pts) If none of the applicants in the first 10 interviews are qualified, what is the conditional probability to meet a qualified applicant in the 11th interview?
(Scratch paper) 9
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12 Forumla Sheet Permutation and Combination: P n k = n! (n k)!, Cn k = ( ) n = k n! k!(n k)!. Laws of probability: Additive law: P (A B) = P (A) + P (B) P (A B); Complement: P (Ā) = 1 P (A), P (B\A) = P (B) P (A B); Conditional Probability: P (A B) = P (A B) ; P (B) Multiplicative law: P (A B) = P (A)P (B A) = P (B)P (A B). Let {B 1,, B k } be a partition of the sample space S such that P (B i ) > 0. Law of total probability: Bayes rule: P (A) = k P (B i )P (A B i ). i=1 P (B j A) = P (B j)p (A B j ) k i=1 P (B i)p (A B i ). Given a discrete random variable Y with probability distribution p(y), The expected value E(Y ) = y yp(y); The variance V (Y ) = E[Y E(Y )] 2 = E(Y 2 ) [E(Y )] 2. Binomial distribution b(n, p): Probability distribution function is p(y) = ( n y) p y (1 p) n y, y = 0, 1,..., n; Expectation = np and Variance = np(1 p). Geometric distribution Geo(p): Probability distribution function is p(y) = (1 p) y 1 p, y = 1, 2, 3,... ; Expectation = 1/p and Variance = 1 p p 2.