Question

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

• A. $\frac{2^n}{5^n}$
• B. $\frac{4^n-2^n}{5^n}$
• C. ${\left(\frac{1}{2}\right)}^n$
• D. $\frac{8^n-4^n}{5^n}$

Answer 1

Answer: (A)

$\frac{2^n}{5^n}$

In any number the last digits can be $0,1,2,3,4,5,6,7,8,9.$
Therefore last digit of each number can be chosen in $10$ ways.
Thus, exhaustive number of ways $={10}^n$.
If the last digit be $1,3,7$ or $9$ none of the number can be even or end in $0$ or $5$.
Thus, we have a choice of $4$ digits $1,3,7$ or $9$ with which each of $n$ numbers sholud end.
So, favorable number of ways $=4^n$.
Hence, the required probability $=\frac{4^n}{{10}^n}=\frac{2^n}{5^n}$