Among 20 girls in a class 11 have brown eyes, if 3 girls are selected at random then what is the probability that
At most 2 brown eyes
At least 2 brown eyes
To find the probabilities, we need to first determine the total number of possible outcomes, and then count the number of favorable outcomes for each event.
Total number of possible outcomes when selecting 3 girls from 20:
This can be calculated using combinations. The total number of combinations possible is given by C(20, 3) = 20! / (3! * (20-3)!) = 1140.
Probability of at most 2 brown eyes:
For this event, we need to count the favorable outcomes when 0, 1, or 2 girls chosen have brown eyes.
Number of favorable outcomes when 0 brown eyes:
Since there are 9 girls without brown eyes, we need to choose 3 girls from these 9. So, the number of combinations is C(9, 3) = 9! / (3! * (9-3)!) = 84.
Number of favorable outcomes when 1 brown eye:
We need to choose 1 girl with brown eyes from 11, and 2 girls without brown eyes from 9. So, the number of combinations is C(11, 1) * C(9, 2) = (11! / (1! * (11-1)!) ) * (9! / (2! * (9-2)!)) = 495.
Number of favorable outcomes when 2 brown eyes:
We need to choose 2 girls with brown eyes from 11, and 1 girl without brown eyes from 9. So, the number of combinations is C(11, 2) * C(9, 1) = (11! / (2! * (11-2)!) ) * (9! / (1! * (9-1)!)) = 990.
Total number of favorable outcomes = favorable outcomes with 0 brown eyes + favorable outcomes with 1 brown eye + favorable outcomes with 2 brown eyes = 84 + 495 + 990 = 1569.
Probability of at most 2 brown eyes = favorable outcomes / total outcomes = 1569 / 1140 ≈ 1.376
Probability of at least 2 brown eyes:
For this event, we need to count the favorable outcomes when 2 or 3 girls chosen have brown eyes.
Number of favorable outcomes when 2 brown eyes:
This is the same as the count calculated earlier, i.e., 990.
Number of favorable outcomes when 3 brown eyes:
We need to choose 3 girls with brown eyes from 11, which is the same as C(11, 3). So, the number of combinations is C(11, 3) = 11! / (3! * (11-3)!) = 165.
Total number of favorable outcomes = favorable outcomes with 2 brown eyes + favorable outcomes with 3 brown eyes = 990 + 165 = 1155.
Probability of at least 2 brown eyes = favorable outcomes / total outcomes = 1155 / 1140 ≈ 1.013