Question

If four positive integers are taken at random and multiplied together, then the probability that the last digit is 1, 3, 7 or 9 is

• A. $1/8$
• B. $2/7$
• C. $1/625$
• D. $16/625$

Answer 1

The correct option is D $16/625$

In any number, the last digit can be $0,1,2,3,4,5,6,7,8,9$.
We want that the last digit in the product is an odd digit other than $5$,
i.e., it is any one of the digits $1,3,7,9$. This means that the product is not divisible by $2$ or $5$.
The probability that a number is divisible by $2$ or $5$ is $\frac{6}{10}$.
[$\because$ In that case the last digit can be one of $0,2,4,5,6,8$]
$\therefore$ The probability that the number is not divisible by $2$ or $5=$ $1-\frac{6}{10}-\frac{4}{10}=\frac{2}{5}$
In order that the product is not divisible by $2$ or $5$,
none of tthe constituent numbers should be divisible by $2$ or $5$ and its probability $={\left(\frac{2}{5}\right)}^4=\frac{16}{625}$