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I have a problem where 5 people put their names in a hat. Each person takes a name out, and does not replace it.

I have these two questions:

what is the probability that a person, call him "Joe" picks his own name out of the hat?

I thought about this, and if he were the first person to pick then it would be 1/5 chance that he picked his own name. But the problem does not state the order in which he picks, so I am kind of confused as to how I would fid the probability. The probability would be different depending on the order with which he pulls a name out correct?

What is the probability that 4 people pick their own name?

I am kind of stuck doing this question. I thought about doing a probability tree where I start off with the two branches of picking your own name, and not picking your own name. But I am not sure this will work.

Any help is appreciated. Thank you.

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First off: the probability 4 people pick their name is 0. Since if 1 person gets it wrong then the name he takes cannot be taken by the one who was supposed to pick that name.

The probability Joe picks his name correctly is $\frac{4!}{5!}$ Since the number of permutations where Joe gets it right are 4! and the total is 5!.

For another approach to the probability Joe gets his name we see:

Probability if he picks:

first: $\frac{1}{5}$

second: $\frac{4}{5}\cdot\frac{1}{4}$

third: $\frac{4}{5}\cdot\frac{3}{4}\cdot\frac{1}{3}$

So really they are all $\frac{1}{5}$

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  • $\begingroup$ Does this help you? Anything unclear? $\endgroup$
    – Asinomás
    Commented Feb 10, 2014 at 22:26
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    $\begingroup$ This helps yes thank you. The only thing I still don't see is how the probability of 4 of them picking themselves is 0. Couldn't theoretically all of them pick themselves? There has to be some chance that happens. $\endgroup$ Commented Feb 10, 2014 at 22:47
  • $\begingroup$ Oh, I see, I thought you meant the probability exactly 4 picked their name, For 5 the probability is $\frac{1}{5!}$ Since only one permutation gives the desired outcome. $\endgroup$
    – Asinomás
    Commented Feb 10, 2014 at 22:48
  • $\begingroup$ Ok thats great thank you. $\endgroup$ Commented Feb 10, 2014 at 22:50

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