We would like to apply the binomial model, so we check the conditions. The number of trials is fixed (\(n = 8\)) (condition 2) and each trial outcome can be classified as a success or failure (condition 3). Because the sample is random, the trials are independent (condition 1) and the probability of a success is the same for each trial (condition 4).

In the outcome of interest, there are \(x = 5\) successes in \(n = 8\) trials (recall that a success is an individual who does *not* exceed the deductible, and the probability of a success is \(p = 0.7\text{.}\) So the probability that 5 of 8 will not exceed the deductible and 3 will exceed the deductible is given by

\begin{align*}
{ 8 \choose 5}(0.7)^5 (1-0.7)^{8-5} \amp = \frac{8!}{5!(5-3)!} (0.7)^5(1-0.7)^{8-5}\\
\amp = \frac{8!}{5!3!} (0.7)^5(0.3)^3
\end{align*}

Dealing with the factorial part:

\begin{gather*}
\frac{8!}{5!3!} = \frac{8\times7\times6\times5\times4\times3\times2\times1} {(5\times4\times3\times2\times1)(3\times2\times1)} = \frac{8\times7\times6}{3\times2\times1} = 56
\end{gather*}

Using \((0.7)^5(0.3)^3 \approx 0.00454\text{,}\) the final probability is about \(56\times 0.00454 \approx 0.254\text{.}\)