# Probability Dice Problem

A pair of dice is tossed 10 times. What is the probability that no 7's or 11's appear as the sum of the sides facing up?

Solution Posted by our instructor:

$\displaystyle P_{\text{sum } 7}=\frac{6}{36}=\frac{1}{6} \\\displaystyle P_{\text{sum } 11}=\frac{2}{36}=\frac{1}{18} \\\displaystyle \sum_{1}^{10} {_nC_r}p^{r}q^{n-r}= \sum_{r=1}^{10} {_{10}C_r}\left(\frac{1}{6}\times\frac{1}{18}\right)^{r}\left(1-\frac{1}{6}\times\frac{1}{18}\right)^{10-r} = 0.08$

I am having trouble understanding this statement - "no 7's or 11's". What i don't understand is why did he let p be the probability that sum of 7 and 11 occurs, since its stated in the problem that no 7 or 11's should appear....?

And why does it need a summation of 1 - 10 successful trials (r) to the binomial equation?

help appreciated

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Your solution is not quite right. I don't even understand what it tries to calculate. – gev Mar 29 '13 at 7:35
If you get the probability of letting the sum to be $7$'s or $11$'s and not getting them as a sum is just a compliment of it. – Inceptio Mar 29 '13 at 7:40

The result is correct, but indeed I do not quite understand the logic of your instructor. I would have said "on a single toss of the dice, the probability that the result is different from 7 or 11 is $1-P_7-P_{11} = 7/9$; since all tosses are independent, the probability it did not happen in 10 tosses is $(7/9)^{10} \approx .081$.

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Actually the instructor's solution evaluates to .08883, instead of your correct .08101. – Byron Schmuland Apr 24 '13 at 21:08

For one try the probability that no 7's or 11's appear as the sum of the sides facing up is $1 - \frac{1}{6} - \frac{1}{18}$.

So for 10 times it is $(1 - \frac{1}{6} - \frac{1}{18})^{10}$.

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