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We roll five six-sided fair dice. a) What is the probability that we get a three dice of one face and two of a second face? (For example, an outcome like 55533.) Explain your counting process. b) What is the probability that the dice all show different faces? Explain your counting process.

a) I worked the problem, but it felt so simple that I don't trust it. For $5$ rolls, you have $6$ options, so the number of possible outcomes is $6^5$. For the first roll, you have $6C1$ choices, for the second and third roll, $1C1$ choice. For the fourth roll you have $5C1$ choices, and then $1C1$ for the fifth roll, Making the probability $(6*5)/6^5$?

b) This one I'm more confident in. Number of possible out comes is still $6^5$, with us having one less option with each roll. $(6*5*4*3*2)/6^5$

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  • $\begingroup$ For (a), you are assuming that the first three dice are the same. This is not usually what is meant by "getting" three dice of one face and two of a second. For example, we want to include cases like 12121 and 11221, not just 11122. Do you see how to count those cases? $\endgroup$ – Matthew Conroy Feb 8 '18 at 4:46
  • $\begingroup$ Three answers with three different results...lol its obvious probability is not a science not a clear one at least lol $\endgroup$ – Isham Feb 8 '18 at 5:41
  • $\begingroup$ Yeah, counting problems always kick my ass because the you can get 3 different answers that are all absolutely correct. $\endgroup$ – Skulloking Feb 8 '18 at 5:58
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For a) think about this: how many pairs there are to make a roll of the kind $aaabb$, that is, how many pairs of $ab$? There are $6\cdot 5=30$ kinds of these pairs. However each throw of the kind $aaabb$ can appear in any order (by example as $ababa$ or $abbaa$), that is, for each pair $ab$ there are $\binom52=10$ different ways that a throw can be ordered.

Thus the total probability is $30\cdot10\cdot \frac1{6^5}$

For the part b) it happen similarly: you want to choose $5$ different numbers of $6$, and these numbers can appear in any order, that is, it can appear as $abcde$ or $acdeb$. How many distinct groups of $5$ numbers, taken from $6$ exists? there are $\binom65=6$, and each one can appear in $5!$ different ways, then the probability is $6\cdot 5!\cdot\frac1{6^5}$.

In part b) you have the same result than me but it is not clear if your reasoning is correct or not.

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You have five dice. Imagine they each have distinctive colours.

Rather than demanding that the 'first' three dice all be the same, count the ways to select three die from five to be the same. Count ways to select one from six numbers for them all to show. Count ways to select one from five numbers for the remaining two die. Put it together. Notice the difference.


For the second problem we can do similarly: Count ways to select 5 distinct numbers from 6. Count ways to arange the dice to show those numbers. Put it together. The answer will match yours, confirming it.

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