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?

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