Question

If a number $x$ is chosen at random from the numbers $2,3,5,7$ and the number $y$ is chosen at random from the numbers $3,4,6,8$, then $P(xy=$ multiple of $3$) ?

• A. $\frac{4}{7}$
• B. $\frac{6}{11}$
• C. $\frac{5}{8}$
• D. $\frac{5}{16}$

Answer 1

The correct option is D $\frac{5}{16}$

Total number of way to choose x and y =4*4=16

For xy to be multiple of 3 either x has to be multiple of 3 or y have to be multiple of 3

So, for x=3 y can be 3,4,6,8

For y=3 x can be 2,3,5,7

For y=6 x can be 2,3,5,7

Total favourable outcomes=4+3+3=10

Probability=10/16=5/8