# A pair of dice rolls and outcome

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?

• Yes, you have done it the right way. Should have said towards the end "$\dots$ getting sum of $\ge 9$ exactly once $\dots$." May 25, 2014 at 13:26
• 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. May 25, 2014 at 13:40

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 $3 \cdot \frac{10}{36} \cdot 1 \cdot \frac{26}{36} = \frac{65}{108}$

• The answer is currently not right. May 25, 2014 at 13:41
• @André Nicolas: Yep, I agree; Now fixed; Thanks May 25, 2014 at 13:50