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Theoretical Probability

Question

A pair of dice is thrown once. The probability that the sum of the outcomes is less than 11 is:
29/36
11/12
7/36
1/6

A

29/36

B

7/36

C

1/6

D

11/12

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Solution

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There are a total of 36 outcomes when two dice are thrown. All the pairs are Favorable outcomes except (6,6), (6,5), and (5,6) whose sum is equal to or more than 11.
Therefore $36 - 3 = 33$ outcomes that are favorable.
Hence,
$P (s u m < 11)$ $=$ $\frac{33}{36} = \frac{11}{12}$

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