Question

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 exceeds 19.
First Roll
Second Roll
B
(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)
The probability of getting two numbers whose sum exceeds 19 is
(Type an integer or a simplified fraction.)

Answer 1

The probability of getting a sum that exceeds 19 from rolling a standard six-sided die twice is 0, as the highest sum possible (6 + 6) is only 12 and cannot exceed 19.

The probability that the sum of two rolls of a die will exceed 19 can only happen if both rolls are 6 (since 6 + 6 = 12 is the highest possible sum and is still less than 19). However, there are no numbers on a standard six-sided die that can be summed to exceed 19. Therefore, there are zero outcomes where the sum exceeds 19. With 36 equally possible outcomes in total when rolling a die twice, the probability can be calculated as follows:

Number of favorable outcomes (sum exceeds 19): 0

Total number of possible outcomes: 36

The probability (P) is therefore P = Number of favorable outcomes / Total number of possible outcomes, which results in P = 0/36 = 0. Hence, the probability is 0, meaning it is impossible to get two numbers whose sum exceeds 19 when rolling a standard six-sided die twice.