Question 1135458: There are 10 girls and 15 boys in a class. If a committee of 5 is selected at random from the class, find the probability that the committee has:
a) Exactly 3 girls;
b) At least one boy;
c) More boys than girls.
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! There are 25C5 = 25!/(5! * (25-5)!) = 53130 ways to select 5 people out of 25
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a) We have 3 girls, then 10C3 = 10!/(3! * (10-3)!) = 120 ways to select 3 girls
:
We have 2 boys, then 15C2 = 15!/(2! * (15-2)!) = 105 ways to select 2 boys
:
Probability(P) of exactly 3 girls chosen = (120 * 105)/53130 = 0.2371 is approximately 0.24
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b) P of at least one boy = 1 - P of all girls
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We have 5 girls, then 10C5 = 10!/(5! * (10-5)!) = 252 ways to select 5 girls
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P of all girls = 252/53130 = 0.0047 is approximately 0.005
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P of at least one boy chosen = 1 - 0.005 = 0.995
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c) more boys than girls implies we have 3 or 4 boys
:
15C3 = 455 and 10C2 = 45, 455 * 45 = 20475 ways to choose 3 boys and 2 girls
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15C4 = 1365 and 10C1 = 10, 1365 * 10 = 13650 ways to chose 4 boys and 1 girl
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P more boys than girls chosen = (20475 + 13650)/53130 = 0.6423 is approximately 0.64
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