On a single roll of two number cubes, what is the probability of rolling doubles given that the sum is 6

1/6

Step-by-step explanation:

We can make a Punnett square to find number of favourable outcome

so

cube1

+

cube2 1 2 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12

The probability of rolling doubles given that the sum is 6

favourable outcome / number of outcome

6 / 36

1 / 6

1/5

Step-by-step explanation:

P (A/B) = P (A and B) / P (B)

Sum of 6: 5/36

(1,5) - - - (5,1) - - - (2,4) - - - (4,2) - - - (3,3)

Probability of rolling doubles: 6/36

(1,1) - - - (2,2) - - - (3,3) - - - (4,4) - - - (5,5) - - - (6,6)

Out of these two "requirements" only one thing overlaps: (3,3)

Therefore:

P (A/B) = (1/36) : (5/36) = 1/5