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I am stuck with an exercise. It says we have a system consisting of two different kinds of components A and B with probabilites to fail of 10% and 20%, respectively. To build the system we use component A 4 times and component B 7 times. Now we are asked to find the probability that at least two components in the system are broken.

I see that we will have to calculate the probability of 0 components failing, as well as the probability of 1 component failing to get to what we want.

$P(0\; components \, fail)= 0.9^4 \cdot 0.8^7 = 0.14 $

However, I do not see how to deal with the fact that there are two different probabilites involved in the case P(1 component failes). Would it be some kind of average like $\frac{1}{2} \left( \binom{4}{1} 0.9^3 \cdot 0.1 \cdot 0.8^7 + \binom{7}{1} 0.9^4 \cdot 0.8^6 \cdot 0.2 \right) $ ?

Thank you very much already for your answers!

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Taking the average is incorrect, but taking the sum is correct. You have correctly calculated the probabilities that exactly one component of either type fails. These two events are disjoint, so their probabilities must simply add to give the correct answer of $$\binom{4}{1} 0.9^3 \cdot 0.1 \cdot 0.8^7 + \binom{7}{1} 0.9^4 \cdot 0.8^6 \cdot 0.2 $$ for the probability that exactly one component fails.

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  • $\begingroup$ Thank you for this fast answer! I actually don't know where I got the idea of averaging from. $\endgroup$ – Moeman Nov 4 '18 at 10:31

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