Probability a product of $n$ randomly chosen numbers from 1-9 is divisible by 10.

Question

I'm working on a problem where each number is chosen randomly from 1-9. Given $n$ numbers chosen in this manner, we multiply all of these together. I'm looking for the probability that this product is divisible by 10.

I reasoned that we only get a multiple of 10 when we multiply 5 and an even number together. So if either all $n$ numbers are odd or if none of the $n$ numbers is 5, then the product will not be divisible by 10, and we can take 1 - P(Not Divisible by 10).

From here, I'm not sure how to come up with a numerical answer when there is an ambiguous number of $n$ randomly chosen numbers. Is there some way to get there? Or is this reasoning flawed?

Answer 1

Not sure this is the most efficient, but:

1. The probability it is not divisible by $5$ is $(1 - 1/9)^{n}$.

2. The probability it is not divisible by $2$ is $(1 - 4/9)^{n}$.

3. The probability it is not divisible by $2$ and $5$ is $(1 - 5/9)^{n}$.

The quantity you seek is $1 - ((1 - 1/9)^{n} +(1 - 4/9)^{n} -(1 - 5/9)^{n})= 1 - (8/9)^n -(5/9)^n + (4/9)^n$.

Answer 2

You choose $n$ numbers. Success occurs when you choose at least one is a five (event $F$) and at least one even number (event $E$). Failure occurs when neither or only one event happens.

You wish to find: $\mathsf P(E\cap F)$

You should the law of complements, the principle of inclusion and exclusion, and that: $$\begin{align} \mathsf P(E^c) &= \frac {5^n}{9^n} \\ \mathsf P(F^c) &= \frac {1}{9^n} \\ \mathsf P(E^c\cap F^c) &= \frac{4^n}{9^n} \end{align}$$