Two dice are thrown at the same time. What is the probability that the sum of the two numbers appearing on the top of the dice is more than 10?
- \(\frac {1}{36}\)
- \(\frac {1}{12}\)
- \(\frac {1}{26}\)
- \(\frac {1}{13}\)
The Correct Option is B
Solution and Explanation
When two dice are thrown, there are a total of $6 \times 6 = 36$ possible outcomes.
We want to find the number of outcomes where the sum of the two numbers is more than 10.
This means the sum is 11 or 12.
The outcomes that result in a sum of 11 are: (5, 6) and (6, 5)
The outcomes that result in a sum of 12 are: (6, 6) So, there are 2 outcomes with a sum of 11 and 1 outcome with a sum of 12. The total number of outcomes where the sum is more than 10 is $2 + 1 = 3$.
Therefore, the probability that the sum of the two numbers is more than 10 is: $$ P(\text{sum} > 10) = \frac{\text{Number of outcomes with sum > 10}}{\text{Total number of outcomes}} = \frac{3}{36} = \frac{1}{12} $$ The probability is $\frac{1}{12}$.
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