Probability of components to fail 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

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
$$
\color{red}{a=\frac1{\frac5{11\color{blue}{t}}+1+\frac{10\color{blue}{t}}6\left(1+\frac{9\color{blue}{t}}7\right)}}.
$$
