08Ma2bHw2Sol

08Ma2bHw2Sol - MA2B HW 2 SOLUTIONS Problem 1 2.1.6 Solution...

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Unformatted text preview: MA2B HW 2 SOLUTIONS Problem 1: 2.1.6 Solution: a. For these independent trials, p = 0 . 7, q = 0 . 3, and according to the binomial distribution, the probability of exactly 4 successes in 8 trials is 8! 4!4! (0 . 7) 4 (0 . 3) 4 = 0 . 1361 b. P (exactly 4 hits | at least 2 hits) = P (exactly 4 hits and at least 2 hits) P (at least 2 hits) = P (exactly 4 hits) P (at least 2 hits) The numerator is 0 . 1361, according to part a . The denominator is equal to 1- P (exactly 0 hits)- P (exactly 1 hit) = 1- (0 . 3) 8- 8(0 . 7)(0 . 3) 7 = 1- . 0000656- . 001224 = 0 . 9987 The answer is then . 1361 . 9987 = 0 . 1363. c. Given that he hits the target with the first two shots, there are 6 remaining independent trials, and the probability that he hits the bull’s eye exactly 4 times in his 8 shots is equal to the probability that he hits exactly 2 times in his 6 unknown shots. The probability of 2 successes in 6 trials is 6! 4!2! (0 . 7) 2 (0 . 3) 4 = 0 . 060 Problem 2: 2.1.12 Solution: a. Recall that in roulette P ( winning ) = p = 18 38 . The event that she stops on the 8th bet is the same as the event where she wins exactly 4 of the first 7 bets, and then also wins the 8th. Since these events are independent, P (stop on 8th) = P (win 4 of 7) P (win) = ( 7! 4!3! p 4 q 3 )( p ) = 0 . 122 b. The event that the gambler has to make 9 or more bets is equal to the probability that she wins 4 or fewer bets in the first 8 trials. This probability is equal to P (0 wins 8 trials) + P (1 win in 8) + · · · + P (4 wins in 8) = q 8 + 8 pq 7 + 8! 6!2! p 2 q 6 + 8! 5!3! p 3 q 5 + 8! 4!4! p 4 q 4 = . 005888 + 0 . 04239 + 0 . 13354 + . 24037 + 0 . 27042 = 0 . 6926 1 2 MA2B HW 2 SOLUTIONS Problem 3: 2.2.9 Solution: a. In this problem, n = 324, p = 0 . 9, μ = np = 291 . 6, and the standard deviation is σ = 324( . 9)( . 1) = 5 . 4. Assuming the passengers show up independently, we can use the normal approximation. The probability that the flight will be overbooked is P (301 ≤ X ≤ 324) ≈ Φ( 324 + 0 . 5- μ σ )- Φ( 301- . 5- μ σ ) = Φ(6 . 1)- Φ(1 . 65) = 1- . 9505 = 0 . 0495 The flight will be overbooked about 1 time in 20. Doing the calculations with the binomial distributions, we see that P ( ≥ 301 showups) = 324 i=301 324 i (0 . 9) i (0 . 1) 324- i = 0 . 0447343 by Mathematica calculations....
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This note was uploaded on 01/08/2011 for the course MA 2b taught by Professor Makarov,n during the Winter '08 term at Caltech.

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08Ma2bHw2Sol - MA2B HW 2 SOLUTIONS Problem 1 2.1.6 Solution...

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