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# Sol V - ENGM2032 Applied Probability and Statistics Winter...

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ENGM2032, Applied Probability and Statistics Winter 2010 Assignment #5: Solutions 1) Let N n be the number of defective rods in a lot of n. Assuming that the rods are independent with a constant probability of being substandard, Bernoulli trials can be used, each rod being a trial, a substandard rod being a ‘success’, and a rod that falls within specs being a ‘failure’, p = 0 . 2 and q = 0 . 8 (note that the ‘success’ here is the exact opposite of what we would normally call a success in the situation in question). Then, N n follows a binomial distribution. a) The expected value (mean) of N 100 is then μ = E N 100 = np = 100(0 . 2) = 20 rods b) Here we are looking for the probability of exactly one ‘success’ in 12 trials, or P N 12 = 1 = 12 1 p 1 q 11 = 12(0 . 2)(0 . 8) 11 = 0 . 2062 . c) P N 10 2 = 1 P N 10 = 0 P N 10 = 1 = 1 10 0 (0 . 2) 0 (0 . 8) 10 10 1 (0 . 2) 1 (0 . 8) 9 = 0 . 6242 . d) For the next two parts, we are counting the number of trials necessary to reach the first success, T 1 . Under the assumptions made, T 1 follows a geometric distribution; hence E T 1 = 1 /p = 5 . (quite naturally, if, on average, you have 2 defective rods in 10, then you expect to test 5 rods, on average, before detecting the first defective one). e) Here we are looking for the probability that we need 5 trials to achieve the first success, P T 1 = 5 = q 4 p = (0 . 8) 4 (0 . 2) = 0 . 08192 . f) Now we are looking for the probability that the 3rd ‘success’ occurs on the 11th trial. This is given to us by the negative binomial distribution, P T 3 = 11 = 10 2 (0 . 2) 3 (0 . 8) 8 = 0 . 0604 . 2) If we assume that unlabelled apples occur independently with constant probability of occur- rence, then apples can be viewed as a sequence of Bernoulli trials with probability of ‘success’ (being unlabelled) of p = 1 / 20 = 0 . 05 and probability of ‘failure’ of q = 0 . 95.

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