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A pair of fair dice is rolled three times and each time the two digits are added. What is the probability that a sum that is greater than or equal to 9 occurs exactly once?

My solution:

The ways to get the various totals is: 9 = 5 4, 4 5 9 = 3 6, 6 3 10 = 5 5 10 = 6 4, 4 6 11 = 5 6, 6 5 12 = 6 6

There are 10 ways to roll to get a sum of >= 9. There are 36 different outcomes for each roll.

The probability of getting a sum of >= 9 is 10/36. The probability of not getting a sum of < 9 is 26/36.

So for 3 rolls for a pair of dice, the probability of getting sum of >= 9 is

$3.(10/36).(26/36).(26/36)$

Correct?

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    $\begingroup$ Yes, you have done it the right way. Should have said towards the end "$\dots$ getting sum of $\ge 9$ exactly once $\dots$." $\endgroup$
    – André Nicolas
    May 25, 2014 at 13:26
  • $\begingroup$ There is a wording issue in your answer, you should have written "the probability that the sum is not $\ge 9$ is $26/36$." But that is what you meant. Please note that your answer of $(3)(10/36)(26/36)^2$ is correct. $\endgroup$
    – André Nicolas
    May 25, 2014 at 13:40

2 Answers 2

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In words:

  • Choose $1$ out of $3$ events for which the sum is greater than or equal to $9$
  • Choose $2$ out of the remaining $2$ events for which the sum is smaller than $9$

Mathematically, it can be expressed as $\binom{3}{1} \cdot P(sum\geq9) \cdot \binom{2}{2} \cdot P(sum<9)$

  • $\binom{3}{1} = \frac{3!}{1!2!} = 3$
  • $\binom{2}{2} = \frac{2!}{2!0!} = 1$
  • $P(sum\geq9) = P([3,6],[4,5],[4,6],[5,4],[5,5],[5,6],[6,3],[6,4],[6,5],[6,6]) = \frac{10}{36}$
  • $P(sum<9) = 1-P(sum\geq9) = \frac{26}{36}$

So the answer is $3 \cdot \frac{10}{36} \cdot 1 \cdot \frac{26}{36} = \frac{65}{108}$

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  • $\begingroup$ The answer is currently not right. $\endgroup$
    – André Nicolas
    May 25, 2014 at 13:41
  • $\begingroup$ @André Nicolas: Yep, I agree; Now fixed; Thanks $\endgroup$
    – barak manos
    May 25, 2014 at 13:50
  • $\begingroup$ @Jaysun: Please note revised answer... $\endgroup$
    – barak manos
    May 25, 2014 at 13:51
  • $\begingroup$ @barak, noted, thank you $\endgroup$
    – Jaysun
    May 25, 2014 at 19:50
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It is actually a Bernoulli trial. and the formula is:nCk*(p^k)*(1-p)^(n-k). 3C1=3

So the answer is 3*(10/36)*(26/36)^2 which is exactly the same as you wrote.

Here is the link to wikipedia "Bernoulli Trial"

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  • $\begingroup$ for a similar question where you wanted a sum greater than or equal to 9 exactly twice, would the answer then be; 3*(10/36)^2(26/36)? $\endgroup$
    – direct1282
    Nov 16, 2020 at 23:20

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