Question

A computer randomly generates an integer from 0 through 9. which expression represents the probability of getting exactly three 0's when randomly choosing 6 numbers?

Answer 1

The probability would be 0.01458.

This is a binomial probability, since the probabilities are independent, there is a fixed number of trials, and there are two outcomes (either a 0 or not a 0).  We use the formula:

$_nC_r(p)^r(1-p)^{n-r}$

The probability of any of the digits being drawn is 1/10.  Then we have:

$_6C_3(0.1)^3(1-0.1)^{6-3} \\ \\_6C_3(0.1)^3(0.9)^3 \\ \\\frac{6!}{3!3!}(0.1)^3(0.9)^3 \\ \\20(0.1)^3(0.9)^3 = 0.01458$