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15B Midterm Math Exam Solutions 26 October 2001 1. Give an example of each of the following or explain why it cannot be done. (a) A bijection from {1, 2, 3, 4} to {a, b, c}. A. Impossible. A bijection has its range and domain the same size. (b) A permutation of {1, 2, 3, 4, 5, 6, 7} that has a cycle of length 4 and also has a cycle of length 5. A. Impossible. The sum of cycle lengths is the size of the set being permuted and 4 + 5 > 7. 2. A committee contains 7 women and 6 men. We want to form a subcommittee with 5 of these people. (a) How many ways can this be done? A. 7+6 5 (b) How many ways can this be done if the subcommittee must contain at least 2 women and at least 2 men? A. There are either (2 men AND 3 women) OR (3 men AND 2 women). By the Rules of Sum and Product, the answer is 6 7 + 6 7 . 2 3 3 2 Choosing two of each sex and then a fth committee member via 6 7 9 over221 counts. For example, if there are 3 women on the committee, the committee is counted 3 times depending on which woman is selected as the fth committee member. 3. The table below gives the joint distribution function, hX,Y , for two random variables X and Y . (a) Find the distribution functions fX for X and fY for Y . A. fX (1) = fX (0) = fX (1) = 1/3. The distribution function for Y is the same. (b) Are X and Y independent? (You must give a correct reason for your answer.) A. No. It suces to give two values r and s for which (X P = r&Y = s) does not equal P (X = r) P (Y = s). Any choices of r and s from {1, 0, 1} have this property. Note: Recall that, if X and Y are independent, then cov(X, Y ) = 0. This table shows that the converse is false: If you do the calculations, you will nd that cov(X, Y ) = 0, but we just showed that X and Y are not independent. hX,Y X=1 X=0 X=+1 Y=1 1/6 0 1/6 Y=0 0 1/3 0 Y=+1 1/6 0 1/6 Math 15B Midterm Exam Solutions 26 October 2001 4. A deck of cards has 52 cards and 13 of these cards are spades. (a) I take seven cards at random from the deck. What is the probability that I get exactly three spades? A. We have a set of size 52 and 13 are bad (spades). The probability that a 3 52 subset of size 7 contains exactly 3 bad is 5213 13 73 7 by the hypergeometric probability formula. (b) I take a card at random from the deck, note whether it is a spade, and put it back. If I do this seven times, what is the probability that I get a spade exactly three times? A. This is a sequence of 7 independent trials. The probability of spade being drawn is 13/52 = 1/4. By the binomial distribution, the probability of drawing exactly 3 spades is 7 (1/4)3 (3/4)4 . 3 You could have left 1/4 and 3/4 as unreduced fractions and switched the roles of spade and non-spade; for example, your answer might have been 7 4 3 4 ((52 13)/52) (1 (52 13)/52) . Suggestion: Think of spade and non-spade like bad and good. END OF EXAM ... View Full Document