Question

A pair of dice is tossed 10 times in succession. What is the probability of getting no 7 and 11 in any of the 10 tosses?

Answer 1

`(28/36)^10~~0.081`

Explanation:

We have 2 fair standard dice. Each die can achieve numbers 1 through 6. And so there are 36 possibilities on each roll.

A 7 can be achieved with: `(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)` = 6 possibilities.

An 11 can be achieved with: `(5,6),(6,5)` = 2 possibilities.

So there are 8 possibilities out of 36 that a 7 or 11 will come up, so the probability is `(8/36)`. This means that the probability of not getting a 7 or 11 is `(1-8/36=28/36)`.

And now let's look at the 10 rolls. Each time we roll the dice, there is a `(28/36)` chance that we won't get a 7 or 11. So over 10 rolls that's `(28/36)` multiplied by itself 10 times, which is `(28/36)^10`, or:

`(28/36)^10~~0.081`