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Out of $120$ people, $5$ need to be selected for a team. What is the probability that you are selected for this team?

$P(A) = \frac{\binom51}{\binom{120}4}$

Is this correct?

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    How do you select 5 from 120? Uniformly without replacement?2012-05-18

2 Answers 2

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No, you want $\frac{\text{number of teams with you on them}}{\text{total number of possible teams}}\;.$ There are $\binom{120}5$ possible teams; how many are there with you on them?

HINT: Imagine that you’ve been chosen. How many ways are there to choose the other four members of the team?

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    @DillPixel: And $\frac{\binom{119}4}{\binom{120}5}=\frac{\frac{119!}{4!115!}}{\frac{120!}{5!115!}}=\frac{119!5!}{120!4!}=\frac5{120}=\frac1{24}$ when all the cancellations have been done. Can you see how the fraction $\frac5{120}$ relates to Jonas Meyer’s answer?2012-05-17
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Brian already gave a good approach, but here is an alternative way to think about it, with another example.

If you randomly eat one third of the cookies on a plate, then the probability that any one particular cookie was eaten is one third. If you rephrased this as saying that there are 54 cookies and 18 of the cookies are going to be chosen to be on a team of eaten cookies, you get a parallel to your problem.

Now that you have it solved, I'll mention that combining the two approaches gives a probabilistic explanation of the identity

$\frac{n-1\choose k-1}{n\choose k}=\frac{k}{n}.$