tauchmeister wrote:
In a group of 10 girls, only 4 have blue eyes. If 3 girls are to be selected at random, without replacement, what is the probability that at least 2 girls will have blue eyes?
A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 2/5
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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