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 $x y$ is more than 16.

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Detailed Solution

Given, a number

is selected at random from the numbers 1, 4, 9, 16 and another number

is selected at random from the numbers 1, 2, 3, 4.
The probability of an event to be occurred is the division of the number of favourable outcomes by the total number of outcomes that is,

Probability of an event = Number of favourable cases Total number of cascs

By choosing the numbers and in the following ways, (1, 1), (1, 2), (1, 3), (1, 4), (4, 1), (4, 2), (4, 3), (4, 4), (9, 1), (9, 2), (9, 3), (9, 4), (16, 1), (16, 2), (16, 3) and (16, 4).
So, the total number of possible outcomes is 16.
Given that the product of numbers is more than 16.
So, the favourable outcomes are (9, 2), (9, 3), (9, 4), (16, 2), (16, 3) and (16, 4).
Thus, the number of favourable outcomes is 6.
Find the required probability by dividing the number of favourable outcomes by the total number of outcomes that is, 6 and 16 respectively.