Out of a group of four men and six women, three people will be randomly chosen to participate in a marketing survey. What is the probability that at least one of the people chosen is a man?
Sorry again, I don't have answer choices
Princeton PS
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- givemeanid
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Let probability that at least one of the people chosen is a man = P(M)
So, the probability that the chosen group contains no man = 1-P(M)
1 - P(M) = Probability that all 3 are women
Probablity that first one chosen is a woman = 6/10
Probablity that second one chosen is a woman after first was a woman = 5/9
Probablity that third one chosen is a woman after first and second were women = 4/8
1 - P(M) = 6/10*5/9*4/8
P(M) = 1 - 120/720 = 1 - 1/6 = 5/6
So, the probability that the chosen group contains no man = 1-P(M)
1 - P(M) = Probability that all 3 are women
Probablity that first one chosen is a woman = 6/10
Probablity that second one chosen is a woman after first was a woman = 5/9
Probablity that third one chosen is a woman after first and second were women = 4/8
1 - P(M) = 6/10*5/9*4/8
P(M) = 1 - 120/720 = 1 - 1/6 = 5/6
So It Goes