Question

Suppose you roll a pair of fair 6-sided dice 36 times. What is the exact probability of getting at least three 9’s?

Answer 1

`((36),(3))(1/4)^3(3/4)^33~~0.0084`

Explanation:

We can find this by using binomial probability:

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

Let's look at the rolls possible in rolling two dice:

`((color(white)(0),ul1,ul2,ul3,ul4,ul5,ul6),(1|,2,3,4,5,6,7),(2|,3,4,5,6,7,8),(3|,4,5,6,7,8,9),(4|,5,6,7,8,9,10),(5|,6,7,8,9,10,11),(6|,7,8,9,10,11,12))`

There are 4 ways to get a 9 out of 36 possibilities, giving `p=9/36=1/4`.

We roll the dice 36 times, giving `n=36`.

We're interested in the probability of getting exactly three 9's, which gives `k=3`

This gives:

`((36),(3))(1/4)^3(3/4)^33`

`((36!)/(33!3!))(1/4)^3(3/4)^33~~0.0084`