Two dice are thrown at the same time. Find the probability that the sum of the two numbers appearing on the top of the dice is more than 10 .

# Two dice are thrown at the same time. Find the probability that the sum of the two numbers appearing on the top of the dice is more than 10 .

1. A
$\frac{11}{12}$
2. B
$\frac{7}{12}$
3. C
$\frac{5}{12}$
4. D
$\frac{1}{12}$

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### Solution:

Given that the two dice are thrown at the same time.
Therefore there are 36 possible outcomes.
The outcomes are,
$\left\{\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(1,4\right),\left(1,5\right),\left(1,6\right),\left(2,1\right),\left(2,2\right),\left(2,3\right),\left(2,4\right),\left(2,5\right),\left(2,6\right),$
$\left(3,1\right),\left(3,2\right),\left(3,3\right),\left(3,4\right),\left(3,5\right),\left(3,6\right),\left(4,1\right),\left(4,2\right),\left(4,3\right)\left(4,4\right)\left(4,5\right),\left(4,6\right),$
$\left(51\right),\left(5,2\right),\left(5,3\right),\left(5,4\right),\left(5,5\right),\left(5,6\right),\left(6,1\right),\left(6,2\right),\left(6,3\right),\left(6,4\right)\left(6,5\right)\left(6,6\right)\right\}$
Total possible outcomes, $n\left(S\right)=36$.
Let E be the event that numbers occurring on top are having the sum as 10 or more than that.
Favourable outcomes are   .  Total number of favourable outcomes, $n\left(E\right)=3$
Probability of event E is,
$P\left(E\right)=\frac{\mathit{Number of favourable outcomes n}\left(E\right)}{\mathit{Total number of outcomes n}\left(S\right)}$
$⇒P\left(E\right)=\frac{3}{36}$
$⇒P\left(E\right)=\frac{1}{12}$
Therefore, the probability of getting two numbers on top whose sum is 10 or more than that is $\frac{1}{12}$.
Hence, option 4 is correct.

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