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Question
Two dice are thrown simultaneously 500 times. Each time the sum of two numbers appearing on their tops is noted and recorded as given in the following table:
Sum | Frequency |
2 | 14 |
3 | 30 |
4 | 42 |
5 | 55 |
6 | 72 |
7 | 75 |
8 | 70 |
9 | 53 |
10 | 46 |
11 | 28 |
12 | 15 |
If the dice are thrown once more, what is the probability of getting a sum more than 10?
Sum
Solution
Total number of times, when two dice are thrown simultaneously, n(S) = 500
Number of times of getting a sum more than 10, n(E1) = 28 + 15 = 43
∴ Probability of getting sum more than 10 = `(n(E_1))/(n(S)) = 43/500` = 0.086
Hence, the probability of getting a sum more than 10 is 0.086
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