Originally Posted by

**pickslides** I like your p value. You should consider the binomial theorem here, it is:

$\displaystyle P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$

In this case $\displaystyle n = 30$, $\displaystyle p = \frac{2}{3}$ and $\displaystyle k = 4$

now substituting in these values we get

$\displaystyle P(X=4) = \binom{30}{4} \left(\frac{2}{3}\right)^4\left(1- \frac{2}{3}\right)^{30-4}$

Can you finish it from here?

This will be the probabilty that 4 devices fail, I think that is what the question is asking?