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?

1 Answer
MathFact-orials.blogspot.com
Oct 8, 2016

#(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#

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