HW2 Sol - Assignment 2 Solutions 1.4.4 Call the events A(having probability 0.1 and B(having probability 0.3 a P A B P A P B b 1 P A B 0.37 c P AB

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Unformatted text preview: Assignment 2 Solutions 1.4.4 Call the events A (having probability 0.1) and B (having probability 0.3). a) P A B P A P B b) 1 P A B 0.37 c) P AB AB P AB 1.4.6 P second spade | first black P first black and second spade P first black P first spade and second spade P first club and second spade P first black 13/52 13/51 13/52 12/51 26/52 25 102 Or you may use a symmetry argument as follows: P second black | first black and P second spade | first black P second club | first black 25/102. 25 . 51 0.9 0.7 0.63 0.1 0.7 0.9 0.3 0.34 P AB by symmetry. Therefore P second spade | first black Discussion. The frequency interpretation is that over the long run, out of every 102 deals yielding a black card first, about 25 will yield a spade second. 1.5.2 a) P W b) P B |W 1.6.3 P WW P B W P W P B W / a) P at least two heads | at least one head P P H H H P at least two H P at least one H 1 2 2 3 3 3 3 1 1 3 0.741 0.963 0.7695 b) P exactly one H| at least one H P at least two H| at least one H 1, because at least one H exactly one H at least two H , and the two events are disjoint. So P exactly one H| at least one H 1 0.7695 0.2305. 2.1.4 P 2 sixes in first five rolls|3 sixes in all eight rolls P 2 sixes in first 5, and 3 sixes in all eight P 3 sixes in all eight P 2 sixes in first five, and 1 six in next three P 3 sixes in all eight 1/6 5/6 1/6 5/6 1/6 5/6 10 3 56 0.535714 2.2.10 We have 30 independent repetitions of a binomial(200, 0.5). For each of these repetitions, the probability of getting exactly 100 heads can be gotten by normal approximation where =(200 * 0.5) = 100 and = 200 0.5 0.5 7.07. The chance of exactly 100 heads can be approximated by P 100 heads 100 0.5 7.07 100 100 0.5 7.07 100 .5282 .4718 .0564 Since the students are independent, the probability that all 30 students do not get exactly 100 heads is 1 P 100 heads and the normal approximation gives us 1 2.2.14 a) # working devices in box has binomial(400, .95) distribution. This is approximately normal with = 380 and 4.36. Required percent 1 . . P 100 heads 1 .0564 0.175 1 2.18 0.0146 (This normal approximation is pretty rough due to the skewness of the distribution. The exact probability is 0.0094 correct to 4 decimal places. The skew normal approximation is 0.0099 which is much better.) b) Using the normal approximation, want largest k so that 1 1.65, so k = 373. 2.2.15 a) x b) x c) Sketch: . . 0.95 so . . x e x x / x x x x x 1 x Outside (4, 4), they are close to zero. d) Let f x . Want f x f x 1 x 1 1 x 1 ...
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This note was uploaded on 05/04/2010 for the course STAT 116 taught by Professor Ross during the Spring '08 term at Stanford.

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