A fair dice is thrown twice What is the probability that the

A fair dice is thrown twice. What is the probability that the total exceeds 7? What is the probability that the total exceeds 7 given that the first thrown is 3?

Solution

2,6 ; 6,2

3,5 ; 5,3

3,6 ; 6,3

4,4

4,5 ; 5,4

4,6 ; 6,4

5,5

5,6 ;6,5

6,6

so total there are 15 options such that sum is more than 7

15 * 1/6 * 1/6 = 15/36

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if the first throw is 3

from the above table we have 3,5 and 3,6 are the only options

so 2 options

the other options which doesnot sum up to exceed 7 are 3,1 ; 3,2 ; 3,3 ; 3,4

so 4 options more

total options are 6 (2+4)

out of them only 2 options has sum more than 7

so answer is 2/6 = 1/3