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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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With a few mathematically-unimportant changes, this is…, so you might have a look at the answers there. – Gerry Myerson May 17 '12 at 23:40
How do you select 5 from 120? Uniformly without replacement? – ablmf May 18 '12 at 1:44
up vote 4 down vote accepted

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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Would it be $$\binom{119}4$$ since one of the spots is taken by me and there's 1 less to choose from? – DillPixel May 17 '12 at 22:38
@DillPixel: Yes, that’s exactly right. – Brian M. Scott May 17 '12 at 22:45
@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? – Brian M. Scott May 17 '12 at 22:50

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

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

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