A single die is rolled twice. the 36 equally-likely outcomes are shown to the right. find the probability of getting two numbers whose sum is 7

1/6

Step-by-step explanation:

When a die is rolled twice the possible outcomes are:

Here the first value represents the first outcome and the second value represents the second outcome.

S = { (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 outcomes = n (S) = 36

Let A be the event that the sum of both outcomes is 7.

A = { (1, 6), (2,5), (3,4), (4,3), (5,2), (6,1) }

n (A) = 6

So, P (A) = n (A) / n (S)

= 6/36

= 1/6