A probability problem by A Former Brilliant Member

The only ways to achieve a sum of $9$ with 5 throws are with permutations of either (i) $3,3,3,B,B$ or (ii) $3,2,2,2,B$ , where $B$ stands for the outcome of a blank face.

In case (i) there are $\dfrac{5!}{3!2!} = 10$ permutations. As the probability of throwing a $3$ is $\dfrac{1}{6}$ and the probability of throwing a $B$ is $\dfrac{4}{6} = \dfrac{2}{3}$ , the probability of obtaining a sum of $9$ in this case is

$10 \times \left(\dfrac{1}{6}\right)^{3} \times \left(\dfrac{2}{3}\right)^{2} = \dfrac{5}{3^{5}}$ .

In case (ii) there are $\dfrac{5!}{3!} = 20$ permutations. Along with the outcome probabilities already noted above, the probability of throwing a $2$ is $\dfrac{1}{6}$ , so the probability of obtaining a sum of $9$ in this case is

$20 \times \dfrac{1}{6} \times \left(\dfrac{1}{6}\right)^{3} \times \dfrac{2}{3} = \dfrac{5}{2 \times 3^{5}}$ .

Thus the total probability of obtaining a sum of $9$ with 5 throws is

$\dfrac{5}{3^{5}} + \dfrac{5}{2 \times 3^{5}} = \dfrac{10 + 5}{2 \times 3^{5}} = \dfrac{5}{162} = \boxed{0.0308}$ to 3 significant figures.

There are two possibilities: 3 times a 3 and 2 blanc or 3 times a 2, a 3 and a blanc, so:

P(sum 9)= {5 \choose 3} \times (\frac{1} {6})^3 \times (\frac{2} {3})^2 + 5\choose 3 \times 3 \choose 1 \times \frac{1} {6})^ 4 \times (\frac{2} {3})=0.031