Question

Two dice numbered 1 to 6 are thrown simultaneously. Find the probability of getting the sum as an odd number greater than or equal to 7.​

Answer 1

Step-by-step explanation:

3/6 that is 1/2 is the final answer

Answer 2

Answer:

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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 =15

Hence, the probability of getting the sum as a prime number = 15/36 = 5/13.

(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 =6

Hence, the probability of getting a total of atleast 10 = 6/36 = 1/6.

(iii)

Favorable outcomes for a doublet of even number are

(2,2),(4,4),(6,6)

Number of favorable outcomes =3

Hence, the probability of getting a doublet of even number = 3/36 = 1/13.

(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 =11

Hence, the probability of getting a multiple of 2 on one dice and a multiple of 3 on the other dice = 11/36.

(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 =12

Hence, the probability of getting a multiple of 3 as the sum = 12/36 = 1/3...

this your answer....