9×29×39×416×216×316×4therefore there are 6 outcomes that are greater than 16.Therefore P = 6/16=3/8.

Problem:
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.

Solution:

To find the probability that the value of xy is more than 16, we need to determine the total number of possible outcomes and the number of favorable outcomes.

Total Number of Outcomes:
Let's first calculate the total number of outcomes for selecting x and y.

- The set of numbers from which x is selected is {1, 4, 9, 16}. So, there are 4 possible outcomes for x.
- The set of numbers from which y is selected is {1, 2, 3, 4}. So, there are 4 possible outcomes for y.

The total number of outcomes is the product of the number of outcomes for x and y, which is 4 * 4 = 16.

Favorable Outcomes:
We need to determine the number of favorable outcomes where the value of xy is more than 16.

To do this, we can list down all the possible combinations of x and y and calculate their product.

x = 1, y = 4, xy = 1 * 4 = 4 (Not more than 16)
x = 1, y = 2, xy = 1 * 2 = 2 (Not more than 16)
x = 1, y = 3, xy = 1 * 3 = 3 (Not more than 16)
x = 1, y = 4, xy = 1 * 4 = 4 (Not more than 16)
x = 4, y = 1, xy = 4 * 1 = 4 (Not more than 16)
x = 4, y = 2, xy = 4 * 2 = 8 (Not more than 16)
x = 4, y = 3, xy = 4 * 3 = 12 (Not more than 16)
x = 4, y = 4, xy = 4 * 4 = 16 (Not more than 16)
x = 9, y = 1, xy = 9 * 1 = 9 (Not more than 16)
x = 9, y = 2, xy = 9 * 2 = 18 (More than 16)
x = 9, y = 3, xy = 9 * 3 = 27 (More than 16)
x = 9, y = 4, xy = 9 * 4 = 36 (More than 16)
x = 16, y = 1, xy = 16 * 1 = 16 (Not more than 16)
x = 16, y = 2, xy = 16 * 2 = 32 (More than 16)
x = 16, y = 3, xy = 16 * 3 = 48 (More than 16)
x = 16, y = 4, xy = 16 * 4 = 64 (More than 16)