Question

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

Answer 1

`~~0.0543`

Explanation:

We can do this using a binomial probability.

`sum_(k=0)^(n)C_(n,k)(p)^k(1-p)^(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=1/3`.

This gives:

`C_(20,10)(1/3)^10(2/3)^(10)`

`184756xx(1/3)^10xx(2/3)^10~~0.0543`