## Problem 3

$\mathrm{n}$ digits, none of them 0 , are randomly (and independently) generated, find the probability that their product is divisible by 10 .


## Solution

Answer: $1-(8 / 9)^{n}-(2 / 9)^{n}+(4 / 9)^{n}$.

A number is divisible by 10 iff it has an even number and a 5 amongst its digits. The probability of no 5 is $(8 / 9)^{\mathrm{n}}$. The probability of no even number is $(5 / 9)^{\mathrm{n}}$. The probability of no 5 and no even number is $(4 / 9)^{\mathrm{n}}$. Hence result.

