Question

If $n$ integers taken at random are multiplied together, the chance (probability) that the last digit of the product is 2,4,6 or 8 is

• A. $\frac{5^n-4^n}{{10}^n}$
• B. $\frac{2^n}{5^n}$
• C. $\frac{4^n-2^n}{5^n}$
• D. None of these

Answer 1

The correct option is C $\frac{4^n-2^n}{5^n}$

It the last digit in product is $2,4,6,8$,
the last digit in all the $n$ numbers should not be $0$ and $5$
and the last digit of all numbers should not be selected exclusively from the set of numbers $\{1,3,7,9\}$
$\therefore$ Favourable number of cases $=8^n-4^n$
but generally the last digits can be only of $0,1,2,\mathrm{3...},9$
Hence, the total number of ways $={10}^n$
Hence, required probablity $=\frac{8^n-4^n}{{10}^n}=\frac{4^n-2^n}{5^n}$