Question

A bowl contains 5 red balls and 5 blue balls. A woman selects 4 balls at random from the bowl. How many different selections are possible if at least 3 balls must be blue?

Answer 1

There are 11 different selections possible if at least 3 balls must be blue.

Step-by-step explanation:

To find the number of different selections, we can consider two cases:

1. Selecting 3 blue balls and 1 red ball
2. Selecting 4 blue balls

Case 1:

The number of ways to select 3 blue balls from 5 is 5 choose 3, which is denoted as C(5, 3) = 10. Then, there is 1 red ball left, so there is only 1 way to select the red ball. Multiplying the number of ways to select the blue balls and the red ball, we get 10 * 1 = 10 possible selections.

Case 2:

Since we need to select all 4 blue balls, there is only 1 way to do so.

Adding the results from both cases, we have a total of 10 + 1 = 11 different selections possible if at least 3 balls must be blue.