Question

If n integers taken at random are multiplied together, show that the chance that the last digit of the product is $1,3,7$, or $9$ is $\frac{2^n}{5^n}$; the chance of its being $2,4,6$, or $8$ is $\frac{4^n-2^n}{5^n}$; of its being $5$ is $\frac{5^n-4n}{{10}^n}$; and of its being $0$ is $\frac{{10}^n-8^n-5^n+4^n}{{10}^n}$.

Answer 1

Each integer can end in $10$ ways

So the total number of integers formed on multiplication $={10}^n$

If the number ends with $1,3,5,7$ then their last digits must be $1,3,7,9$

So total number of such numbers $=4^n$

Thus probability of selecting such number $=\frac{4^n}{{10}^n}=\frac{2^n}{5^n}$

If number ends with $2,4,6,8$ then the last digit must not be $0,5$ and the numbers of the last case are also excluded

Thus total number of such numbers $8^n-4^n$

Probabilty of selecting such numbers $=\frac{8^n-4^n}{{10}^n}=\frac{4^n-2^n}{5^n}$

If number ends with $5$ then digits must be $1,3,5,7,9$ excluding the number formed by $1,3,7,9$

So total number of such numbers $=5^n-4^n$

Probability of selecting such numbers $=\frac{5^n-4^n}{{10}^n}$

Probability of selecting number that end with $0$ $=1-$number in the above cases

$=1-\frac{4^n}{{10}^n}-\frac{8^n-4^n}{{10}^n}-\frac{5^n-4^n}{{10}^n}=\frac{{10}^n-5^n-8^n+4^n}{{10}^n}$

Hence proved