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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
- 3?
- more than 10?
- less than or equal to 5?
- between 8 and 12?
Solution
Total number of times, when two dice are thrown simultaneously, n(S) = 500
i. Number of times of getting a sum 3,
n(E) = 30
∴ Probability of getting a sum 3 = `(n(E))/(n(S))`
= `30/500`
= `3/50`
= 0.06
Hence, the probability of getting a sum 3 is 0.06
ii. 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
iii. Number of times of getting a sum less than or equal to 5,
n(E2) = 55 + 42 + 30 + 14 = 141
∴ Probability of getting a sum less than or equal to 5 = `(n(E_2))/(n(S))`
= `141/500`
= 0.282
Hence, the probability of getting a sum less than or equal to 5 is 0.282.
iv. The number of times of getting a sum between 8 and 12,
n(E3) = 53 + 46 + 28 = 127
∴ Required probability = `(n(E_3))/(n(S))`
= `127/500`
= 0.254
Hence, the probability of getting a sum between 8 and 12 is 0.254.
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