There is a group of six people; among them are Mr X and Mr Y. Three persons are chosen at random without repetitions. Find the probability of choosing Mr X, but not Mr Y.
I'd start by figuring out the probability of choosing Mr. X -- he's 1 out of 6, and you pick 3 people, so what's the probability of picking Mr. X?
When the first person is chosen there is a 1 out of 6 chance of choosing Mr X or Mr Y. When the second person is chosen there is a 1 out of 5 chance of choosing Mr X or Mr Y. When the third person is chosen there is a 1 out of 4 chance of choosing Mr X or Mr Y. Therefore the chance of choosing both Mr X and Mr Y is the product of the separate probabilities: \[P(X+Y)=\frac{1}{6}\times \frac{1}{5}\times \frac{1}{4}=\frac{1}{120}\]
@kropot72: This question asked for "Fnd the probability of choosing Mr X, *but not* Mr Y. "
You are quite right FoolForMath. So i shall try again: When choosing the first person there is a 1 out of 6 chance of choosing Mr X. When choosing the second person there is a 4 out of 5 chance of not choosing Mr Y. When choosing the third person there is a 3 out of 4 chance of not choosing Mr Y. Therefore the probability of choosing Mr X and not choosing Mr Y is the product of the separate probabilities of the sub-events: \[P(XnotY)=\frac{1}{6}\times \frac{4}{5}\times \frac{3}{4}=\frac{12}{120}\]
I think it may be a bit simpler, you can do it in just two steps. What is the probability of choosing Mr. X? He's 1 of 6, and you draw three times. That probability indicates the times when Mr. X is one of the three "in" slots, leaving two "in" slots and three "out" slots. In that situation, what is the probability that Mr. Y will be in one of the "out" slots? Multiply those probabilities together (the odds of X being in, and then, given that condition, the odds of Y being out) and you've got your answer. @kropot72, I think your formula was missing the fact that on the second draw, if Mr. X wasn't picked the first time, he was still available to be picked.
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