1. ## Sample size

Hi
I don't understand how to approach this question, could someone give me a hint.

Thank you

2. ## Re: Sample size

Each trial is taking a shot, and in each trial there are only two possibilities - hit or miss. So that means the distribution is Binomial. So p = 0.6 but n is unknown.

We want to evaluate n so that \displaystyle \begin{align*} \textrm{Pr}\,\left( X = 5 \right) = 0.25 \end{align*}, so

\displaystyle \begin{align*} {n\choose{5}} \left( \frac{3}{5} \right) ^5 \left( \frac{2}{5} \right) ^{n - 5} &= \frac{1}{4} \\ \frac{n!}{5! \left( n - 5 \right) !} \left( \frac{3}{5} \right) ^5 \left( \frac{2}{5} \right) ^{n - 5} &= \frac{1}{4} \end{align*}

I doubt this can be solved by hand, you will need to use a CAS.