# Bill rolls a fair number cube 20 times. What is the probability that 10 of his results are multiples of 3?

$\approx 0.0543$

#### Explanation:

We can do this using a binomial probability.

${\sum}_{k = 0}^{n} {C}_{n , k} {\left(p\right)}^{k} {\left(1 - p\right)}^{n - k} = 1$

For this question, $n = 20 , k = 10$

With a roll of a fair number cube, see that the multiples of 3 are 3 and 6, giving a probability of rolling a multiple of 3 on any given roll as $p = \frac{1}{3}$.

This gives:

${C}_{20 , 10} {\left(\frac{1}{3}\right)}^{10} {\left(\frac{2}{3}\right)}^{10}$

$184756 \times {\left(\frac{1}{3}\right)}^{10} \times {\left(\frac{2}{3}\right)}^{10} \approx 0.0543$