Question

A number x is selected at random from the numbers 1, 4, 9, 16 and another number y is selected at random from the numbers 1, 2, 3, 4. Find the probability that the value of xy is more than 16.

Answer 1

Let x be 1, 4, 9, 16 and y be 1, 2, 3, 4.

Now, $xy=\{1,2,3,4,4,8,12,16,9,18,27,36,16,32,48,64\}$

Total number of possible outcomes =16

Number of outcomes where product is more than $16=6$

i.e., $\{18,27,36,32,48,64\}$

$\therefore$ Required probability = $\frac{6}{16}=\frac{3}{8}$