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I roll a die repeatedly. Find the chance that the first 4 rolls all show different faces, and the 5th roll shows a face that has appeared before.

$P($First 4 rolls shows different faces$)=\frac{5}{6}\times \frac{4}{5}\times \frac{3}{4}\times \frac{2}{3}=?$

$P($5th face appeared before$)=\frac{5}{6}?$

Product Rule= $P($First 4 roles$)\times P($5th Face$)=(\frac{5}{6})^2\times \frac{4}{5}\times \frac{3}{4}\times \frac{2}{3}=?$

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  • $\begingroup$ Parentheses, please. You have many divides and multiplies, please make sure we know what is on top and what is on the bottom. $\endgroup$ Apr 30, 2013 at 4:42
  • $\begingroup$ IMO, computing probabilities is an exercise in organization moreso than anything else. Explain what you trying to do in your calculation. It may help to do one step at a time rather than all at once. $\endgroup$
    – user14972
    Apr 30, 2013 at 4:47

3 Answers 3

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Hints:

First, how many outcomes exist to get four different numbers?

  1. The first roll can have $6$ different outcomes, the second one should have $5$ and the third, and fourth?

  2. What is the size of the sample space? (each time there exists $6$ possibilities.)

  3. The fifth roll should be among those first four rolls so has $4$ possible outcomes so the probability is $4/6.$

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  • $\begingroup$ This helped OP, so +1. $\endgroup$ Apr 30, 2013 at 5:22
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(1) * (5/6) * (4/6) * (3/6) * (4/6)

The probability on the first roll is any roll, so it's one.

The probability on the second roll is the probability of it being any 1 of the 5 not yet seen faces, which is 5* (1/6)

The probability on the third roll is the probability of it being any 1 of the 4 not yet seen faces, which is 4*(1/6)

The probability on the fourth roll is the probability of it being any 1 of the 3 not yet seen faces, which is 3*(1/6)

The probability on the fith roll is the probability of it being any of the 4 already seen faces, which is 4*(1/6)

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  • $\begingroup$ Jatorious understood! $\endgroup$ Apr 30, 2013 at 5:06
  • $\begingroup$ I think this helped OP, so +1. Sorry you didn't get the checkmark. It happens. $\endgroup$ Apr 30, 2013 at 5:21
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Hint: for two rolls, the chance is $\frac 56$ that the second is not the same as the first. Given that, what is the chance that the third roll is different from either of the first two?

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  • $\begingroup$ Explain the conditional probability scheme hinted in Principle? $\endgroup$ Apr 30, 2013 at 4:55
  • $\begingroup$ Thanks Ross Milikan got it! $\endgroup$ Apr 30, 2013 at 5:04

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