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 \\ \\(6!)/(3!3!)(0.1)^3(0.9)^3 \\ \\20(0.1)^3(0.9)^3 = 0.01458`