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Kindly help me to answer this:

Suppose five integers are chosen successively at random between 0 and 13, inclusive. Find the probability that:

(Round your answer to four decimal places if necessary.)

a. they are all different

b. not more than 2 are the same?

Sebastian Nielsen
  • 612
Michael Leuterio
  • 11

1 Answers1

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The first can be anything. The chance the second does not match the first is $\frac {13}{14}$, then the chance the third does not match either is $\frac {13}{14}$ and so on, giving a chance of no match as $\frac {13\cdot 12 \cdot 11 \cdot 10}{14^4}$

For the second, compute the chance of exactly one match then add to the first. You can choose the locations of the match in ${5 \choose 2}=10$ ways. You have $\frac 1{14}$ chance that those locations match, then $\frac {13\cdot 12 \cdot 11 }{14^3}$ chance the other three are different from the one at those locations and each other. The chance comes out the same, so you can just double the first result.

Ross Millikan
  • 374,822
  • The second question asks for the probability that not more than two are the same, so you have to add the two probabilities you found. – N. F. Taussig Oct 04 '18 at 14:10
  • @N.F.Taussig: I missed that. We can just add my second to the first. Updated. Thanks – Ross Millikan Oct 04 '18 at 14:25
  • Can you detail the solution on letter B? Thank you so much – Michael Leuterio Oct 05 '18 at 06:58
  • Say you choose the first and third that are the two that match. There are $10$ ways to choose the pair. Then the first can be anything and you have $\frac 1{14}$ chance that the third matches. Then you have to fill in the other three slots with numbers that do not match that one or each other. – Ross Millikan Oct 05 '18 at 13:57