Question

If $n$ integers taken at random are multiplied together, then the probability that the last digit of the product is $1,3,7$ or $9$ is:

• A. $\frac{2^n}{{10}^n}$
• B. $\frac{2^n}{5^n}$
• C. $\frac{5^n}{{10}^n}$
• D. $\frac{4^n-2^n}{{10}^n}$

Answer 1

The correct option is B $\frac{2^n}{5^n}$

Last digit can be $0,1,2,3,4,5,6,7,8,9$
Therefore, last digit of each number can be chosed in 10 ways
Thus, exhaustive number of ways $={10}^n$
If the last digit be $1,3,7$ or $9$, then none of the number can be even or end with $0$or$5$
Thus, we have choice of $4$ digits $1,3,7$or$9$ with which each of $n$ number should end.
So, favorable number of ways $=4^n$
Required probability $=\frac{4^n}{{10}^n}$