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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. – André Nicolas Apr 28 '12 at 21:37
@AndréNicolas $P(5 \cap \{4,5,6,7\}) = P(5)$ right? – Andrew Apr 28 '12 at 21:38
Yes, that's right. – David Mitra Apr 28 '12 at 21:39
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. – André Nicolas Apr 28 '12 at 21:40

You already know the answer is $a=p_5/(p_4+p_5+p_6+p_7)$ where $p_k=\mathrm P(X=k)$. Further simplifications occur if one considers the ratios $r_k=p_{k+1}/p_k$ of successive weights. To wit, $$ r_k=\frac{{n\choose k+1}p^{k+1}(1-p)^{n-k-1}}{{n\choose k}p^{k}(1-p)^{n-k}}=\frac{n-k}{k+1}\color{blue}{t}\quad\text{with}\ \color{blue}{t=\frac{p}{1-p}}. $$ Thus, $$ \frac1a=\frac{p_4}{p_5}+1+\frac{p_6}{p_5}+\frac{p_7}{p_5}=\frac1{r_4}+1+r_5(1+r_6), $$ which, for $n=15$ and with $\color{blue}{t=\frac3{17}}$, yields $$ \color{red}{a=\frac1{\frac5{11\color{blue}{t}}+1+\frac{10\color{blue}{t}}6\left(1+\frac{9\color{blue}{t}}7\right)}}. $$

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