Question

If $4$ integers are taken randomly and multiplied together then the probability that the last digit of the product is among 1, 3, 7 or 9

• A. ${\left(\frac{3}{5}\right)}^4$
• B. ${\left(\frac{2}{5}\right)}^4$
• C. ${\left(\frac{1}{5}\right)}^4$
• D. None of these

Answer 1

The correct option is B ${\left(\frac{2}{5}\right)}^4$

There are $10$ digits $0,1,2,...,9$ any of which can occur in any number at the last place, i.e., at the unit place.

It is obvious that if the last digit in any of the four numbers is $0,2,4,5,6,8$ then the product of any of such four numbers will not give a number having its last digit as $1,3,7,9$. Hence, it is necessary that the last digit in each of the four numbers must be any of the four digits $1,3,7,9$

Thus for each of the four numbers, the number of ways for the last digit $=10$ and favourable numbers of ways for the last digit $=4$

$\therefore$ the probability that the last digit is any of the four numbers $1,3,7,9$ is $=\frac{4}{10}=\frac{2}{5}$

Hence, the required probability that the last digit in each of the four numbers is $1,3,7,9$ so that the last digit in their product is $1,3,7,9={\left(\frac{2}{5}\right)}^4=\frac{16}{625}$