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प्रश्न
In a single throw of two dice, find the probability of a total of at most 10.
उत्तर
The number of possible outcomes is 6 × 6 = 36.
We write them as given below:
1, 11, 21, 31, 41, 51, 6
2, 12, 22, 32, 42, 52, 6
3, 13, 23, 33, 43, 53, 6
4, 14, 24, 34, 44, 54, 6
5, 15, 25, 35, 45, 55, 6
6, 16, 26, 36, 46, 56, 6
n(S) = 36
E = Event of getting a total of at most 10
= {(1, 1),(1, 2),(1, 3),(1, 4),(1, 5),(1, 6),(2, 1),(2, 2),(2, 3),(2, 4),(2, 5),(2, 6),(3, 1),(3, 2),(3, 3),(3, 4),(3, 5),(3, 6),(4, 1),(4, 2),(4, 3),(4, 4),(4, 5),(4, 6),(5, 1),(5, 2),(5, 3),(5, 4),(5, 5),(6, 1),(6, 2),(6, 3),(6, 4)}
Therefore total number of favorable ways = 33 = n(E)
Probability of getting a total of at most 10 = `(n(E))/(n(s)) = 33/36 = 11/12`
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