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I want to verify my reasoning with you.

An electronic system contains 15 components. The probability that a component might fail is 0.15 given that they fail independently. Knowing that at least 4 and at most 7 failed, what is the probability that exactly 5 failed?

My solution:
$X \sim Binomial(n=15, p=0.15)$
I guess what I have to calculate is $P(X=5 | 4 \le X \le 7) = \frac{P(5 \cap \{4,5,6,7\})}{P(\{4,5,6,7\})}$. Is it correct? Thank you

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    Yes, the approach is correct.2012-04-28
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    @AndréNicolas $P(5 \cap \{4,5,6,7\}) = P(5)$ right?2012-04-28
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    Yes, that's right.2012-04-28
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    Yes, of course. This sort of thing happens a lot in conditional probabilities, there is less work to do than it seems at first sight.2012-04-28

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