STAT 263 Lecture 4: Homework Solutions 4 - Chapters 6
15 Aug 2015
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P(at most 3 mistakes) = 0. 64 which could also be expressed as p(x 3) P(4 to 6 mistakes) = 0. 21 which could also be expressed as p(4 x 6) Remember that the sum of all the probabilities of x being 0, 1, 2, 3, 4, is 1. You could think of the probability of making at most 3 mistakes as the probability of making 0 or 1 or. And the probability of making 4 to 6 mistakes as the probability of making 4 or 5 or 6 mistakes. So if you add these two probabilities together, you get the probability of making 0 or 1 or 2 or 3 or 4 or 5 or 6 mistakes, which is exactly what part b) asks you to find. In parts a) and c) you need to find the complement of a probability you are given (a)) or have just found (c))
