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Consider a system of components in which there are 4 independent components, each of which possesses an operational probability of 0.9. The system does have a redundancy built in such that it does not fail if 3 out of the 4 components are operational. What is the probability that the total system is operational?

I have no idea how to approach this. Can someone help?

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    **Hint.** Binomial distribution. – 2017-02-08
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    So 4 independent trials and probability of success 0.9. It is Binomial. So if X represents the number of components operational, do I need to calculate P(X=3)? – 2017-02-08

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The system fails when two, or more, components fail. If each component falls out independently (which is probably true) then we have a binomial distribution $Bin(0.9)$. Now it is easy to calculate the probability of system failure. $$P( \text{failure}) = 1 - P(\text{operational}) = 1 - \left(P(\text{#fallout} = 0)+ P(\text{#fallout} = 1)\right)$$

$$ = 1-\left(\binom{4}{0}0.9^4 + \binom{4}{1}0.9^3\cdot0.1\right)$$

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    read more closely ... "does not fail if 3 out of 4 are operational". the system fails when two components fail – 2017-02-08
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    I corrected my mistake. – 2017-02-08
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    The naked probability that exactly 2 of the components fail is not really what is asked for here, because the system will be just as unavailable if exactly 3 or exactly 4 components fail. – 2017-02-08
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    It is a binomial distribution of 4 independent trials each with a probability of success 0.9. So we are asked to find P(X=3)+P(X=4) where X= number of components working. Since at least 3 must work for the system to be functional, hence the summation. Apply the PMF of Binomial distribution to get the answer. – 2017-02-08
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    Trying to answer so quickly I don't even read. Now I understand – 2017-02-08