A committee of 5 people is to be selected from student council. Council has 6 boys and 7 girls. What is the probability that the committee will have at least 3 boys?
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To find the probability that the committee will have at least 3 boys, we can break down the problem into several cases: selecting 3 boys and 2 girls, selecting 4 boys and 1 girl, or selecting 5 boys and no girls.
Case 1: Selecting 3 boys and 2 girls
The number of ways to choose 3 boys from 6 is given by the combination formula: C(6, 3) = 6! / (3! * (6-3)!) = 20
Similarly, the number of ways to choose 2 girls from 7 is given by: C(7, 2) = 7! / (2! * (7-2)!) = 21
The total number of ways to select 5 people from the council is: C(13, 5) = 13! / (5! * (13-5)!) = 1287
Therefore, the probability of selecting 3 boys and 2 girls is: (20 * 21) / 1287 ≈ 0.311
Case 2: Selecting 4 boys and 1 girl
The number of ways to choose 4 boys from 6 is given by: C(6, 4) = 6! / (4! * (6-4)!) = 15
Similarly, the number of ways to choose 1 girl from 7 is given by: C(7, 1) = 7! / (1! * (7-1)!) = 7
The total number of ways to select 5 people from the council is: C(13, 5) = 1287 (as calculated before)
Therefore, the probability of selecting 4 boys and 1 girl is: (15 * 7) / 1287 ≈ 0.090
Case 3: Selecting 5 boys and no girls
The number of ways to choose 5 boys from 6 is given by: C(6, 5) = 6! / (5! * (6-5)!) = 6
The total number of ways to select 5 people from the council is: C(13, 5) = 1287 (as calculated before)
Therefore, the probability of selecting 5 boys and no girls is: 6 / 1287 ≈ 0.005
To find the probability of at least 3 boys, we just need to sum up the individual probabilities from the three cases:
0.311 + 0.090 + 0.005 ≈ 0.406
So, the probability that the committee will have at least 3 boys is approximately 0.406, or 40.6%.