Asingle 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

24/56=hx^7

step-by-step explanation:

i believe the answer is the polygon, heptagon.