A weather reporter for a channel is forecasting a $30\%$ chance of rain for today and the next four days. Find the probability of having rain on exactly one day.
Let $p$ be the probability of raining on a given day.
Let $q$ be the probability of not raining on a given day.
Given: $\;$ $p = 30\% = 0.3$; $\;$ $q = 1- p = 1 - 0.3 = 0.7$
Total number of days $= 4$
$\begin{aligned}
P \left(\text{It rains exactly one day}\right) & = {^{4}}{C}_{1} \times \left(p\right)^{1} \times \left(q\right)^{4 - 1} \\\\
& = \dfrac{4!}{1! \times 3!} \times 0.3 \times \left(0.7\right)^{3} \\\\
& = 4 \times 0.3 \times \left(0.7\right)^{3} \\\\
& = 0.4116 \approx 41.2\%
\end{aligned}$