Question

# Two dice are thrown simultaneously. Find the probability of getting: (i)The sum as a prime number. (ii) A total of at least 10. (iii) A doublet of even number. (iv) A multiple of 2 on on. dice and a multiple of 3 on the other dice. (v) A multiple of 3 am the sum.

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Solution

## Sample space for total number of possible outcomes(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),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)Total number of outcomes =36(i)Favorable outcomes for sum as prime are(1,1),(1,2),(1,4),(1,6),(2,3),(2,3),(2,5),(3,2),(3,4),(4,1),(4,3),(5,2),(5,6),(6,1),(6,5)Number of favorable outcomes =15Hence, the probability of getting the sum as a prime number. = 1536=512(ii)Favorable outcomes for total of atleast 10 are(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)Number of favorable outcomes =6Hence, the probability of getting a total of atleast 10 = 636=16(iii)Favorable outcomes for a doublet of even number are(2,2),(4,4),(6,6)Number of favorable outcomes =3Hence, the probability of getting a doublet of even number = 336=113(iv)Favorable outcomes for a multiple of 2 on one dice and a multiple of 3 on the other dice are(2,3),(2,6),(3,2),(3,4),(3,6),(4,3),(4,6),(6,2),(6,3),(6,4),(6,6)Number of favorable outcomes =11Hence, the probability of getting a multiple of 2 on one dice and a multiple of 3 on the other dice = 1136(v)Favorable outcomes for getting a multiple of 3 as the sum(1,2),(1,5),(2,1),(2,4),(3,3),(3,6),(4,2),(4,5),(5,1),(5,4),(6,3)(6,6)Number of favorable outcomes =12Hence, the probability of getting a multiple of 3 as the sum = 1236=13

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