Probability when given the percentage chance There is a 30% chance that a driver will have an accident in their first year of driving. From 18 people getting their license this year in June, what is the probability that less than a quarter will have an accident before June next year?
So i had a go at it. 30/100 *18= 5.4 people have a probability of an accident. Less than 1/4 of 18 is about 4.5 people. So if 30% of 18 is 5.4 then 4.5 will be 0.3*4.5/5.4= 0.25 chance, but this is incorrect. Any ideas?

Assuming that the events are independent, this is equal to a binomial distribution. Calculate the chance that less than $\frac14$ of the 18 people have an accident is the same as calculating if less than or equal to 4 people have an accident. We can use a table for the cumulative distribution function, which would give us 0.333 or we could calculate it using the probability mass function
$P(X = x) = f(x) = {n \choose x}p^x(1-p)^{n-x}$
$\begin{align*}P(X \le 4) &= P(X = 0) + P(X=1) + P(X=2) + P(X=3) + P(X = 4) \\
&= {18 \choose 0} 0.7^{18} + {18 \choose 1} 0.3\cdot0.7^{17} + {18 \choose 2}0.3^2\cdot 0.7^{16} + {18 \choose 3}0.3^3 \cdot 0.7^{15} + {18 \choose 4}0.3^4\cdot 0.7^{14} \\
&\approx 0.3327
\end{align*}$
