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