Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
1  
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
1  
Yes, that's right. –  David Mitra Apr 28 '12 at 21:39
3  
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
add comment

1 Answer

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.