In a group of 10 girls, only 4 have blue eyes. If 3 girls ar

Total selection = 10C3 = 10 * 3 * 4
Exactly 1 girl has blue eyes = 4C1 = 4
Exactly 3 girls don't have blue eyes = 6C3 = 5 * 4

At least 1 girl has blue eyes = \(1 - [(4 * 5 * 4)/ 10 * 3 * 4]\) = \(1 - 2/3\) =\(1/3\)
Ans.: B

We are given that we have a group of 10 girls; 4 girls have blue eyes and 6 girls do not have blue eyes. We need to determine the probability, when 3 girls are selected, that at least 2 girls will have blue eyes. There are two scenarios in which we can select at least 2 girls with blue eyes.

Scenario 1: Blue-Blue-Not Blue

Scenario 2: Blue-Blue-Blue

Before calculating each scenario, let’s first determine the total number of ways to select 3 girls from a group of 10 girls.

The number of ways to select 3 girls from a group of 10 is:

10C3 = (10 x 9 x 8)/3! = (10 x 9 x 8)/(3 x 2 x 1) = 5 x 3 x 8 = 120

Next we need to determine the number of ways to select the blue-eyed girls.

First, let’s calculate scenario 1: Blue-Blue-Not Blue

The number of ways to select 2 girls with blue eyes from 4 is:

4C2 = 4!/2!2! = (4 x 3)/2! = 6 ways

The number of ways to select 1 girl with no blue eyes from 6 is:

6C1 = 6 ways

Thus, the probability of selecting 2 girls with blue eyes and 1 with no blue eyes is:

(6 x 6)/120 = 36/120

Next, let’s calculate scenario 2: Blue-Blue-Blue

The number of ways to select 3 girls with blue eyes from 4 is:

4C3 = (4 x 3 x 2)/3! = 4 ways

Thus, the probability of selecting 3 girls with blue eyes is:

4/120

Finally, we can determine the probability of selecting at least 2 girls with blue eyes:

36/120 + 4/120 = 40/120 = 1/3

Answer: B

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