Determine the expected value, variance, and standard deviation of X.?

X =?(4*0.12*0.88) = 0.65 In general, for any binomial distribution P(X = 0) = (1-?)^n In this case P(X = 0) = 0.88^4 = 0.60 M pls check again your computation. For n = 4, k = 0 and p = 0.12 the binomial formula n! /k!

*(n - k)! * p^k * (1 - p)^(n - k) gives 4! /0!

*4! * 0.12^0 * 0.88^4 = 1*1*0.88^4 = 0.60, not 1. Unless you really meant that p = 0.12 is the same as p = 0.12%, and put 0.0012 into your formula and then rounded it off ...

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