Question

Four whole numbers taken at random are multiplied together. What is the chance that the last digit in the product is $1,3,7$ or $9?$

• A. $\frac{16}{625}$
• B. $\frac{1}{210}$
• C. $\frac{8}{125}$
• D. $\frac{4}{25}$

Answer 1

Answer: (A)

$\frac{16}{625}$

If the product of the four numbers ends in one of the digitd $1,3,7$ or $9$, each number should have the last digit as one of these $4$ digits.
$\therefore$ the number of favorable cases $=4^4$
Total number of all possible cases ${10}^4$
Hence, the required probability $=\frac{4^4}{{\left(10\right)}^4}=\frac{2^4}{5^4}=\frac{16}{625}$