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5 dice are rolled. What is the probability of getting 2 sixes and 2 one's.

My attempt: $\binom{5}{2}\times\binom{3}{2}\times(\frac{1}{6})^4\times(\frac{4}{6})$ Is this correct?

Thanks

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  • $\begingroup$ What does $C$ denote? $\endgroup$ – JohnWO Sep 29 '14 at 13:14
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lgtm.

your answer would look "prettier" (more symmetric) if you used the multinomial coefficient $\binom{5}{2,2,1}$ instead of the product of binomials.

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  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. $\endgroup$ – Travis Willse Sep 29 '14 at 13:35
  • $\begingroup$ @Travis: the question is "Is this correct". I did answer that: "lgtm==yes". $\endgroup$ – sds Sep 29 '14 at 13:37
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Yeah you are correct.

  1. The total number of ways the 5 dices shows up is $6^5$.

  2. The number of ways we can arrange two 6's , two 1's and one number I.e.,(6, 6, 1, 1, x) where x can be 2 or 3 or 4 or 5 but not 6, 1

is $5!/(2! 2!)$

  1. Choosing the number x is again 4 ways

So the answer is $4\cdot(5!/(2! 2!))/6^5$

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  • $\begingroup$ what is wrong with my answer? $\endgroup$ – user108605 Sep 29 '14 at 13:35
  • $\begingroup$ I think you are wrong by a factor of 4 - the number of ways you can chose the value of $x$. $\endgroup$ – sds Sep 29 '14 at 13:37
  • $\begingroup$ Yeah @sds. Edited now... you are right with your answer user108605 $\endgroup$ – Jasser Sep 29 '14 at 13:41
  • $\begingroup$ Thankyou for pointing out my mistake @sds $\endgroup$ – Jasser Sep 29 '14 at 13:44
  • $\begingroup$ @user291957: you are welcome! $\endgroup$ – sds Sep 29 '14 at 13:45

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