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#1
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Probability
If we have 5 machines, the probability that any particular machine fails is 20% or P(A) = 0.2. So, P(A)=0.2, P(B)=0.2, P(C)=0.2, P(D)=0.2, P(E)=0.2. I need to know what the probability of any one of the events occuring. More than one machine can fail at once.
I tried this: P(P(P(P(AorB)orC)orD)orE) = 0.67232 |
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#2
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This is a binomial probability. Let X be the number of failures. X is distributed as binomial with n=5, p=0.20.
P(more than one machine can fail at once) = P(X>=2) =1-P(X=0)-P(X=1) You can use binomial formula to get P(X=0) and P(X=1). |
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