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An ordinary (fair) die is a cube with the numbers 1 through 6 on the sides (represented by painted spots). Imagine that such a die is rolled twice in successionand that the face values of the two rolls are added together. This sum is recorded as the outcome of a single trial of a random experimentCompute the probability of each of the following events.Event A: The sum is greater than 6.Event B: The sum is divisible by 2.Write your answers as exact fractions.DO(a) P(A) = 0Х5?(b) P(B) =

Sagot :

The sum is greater than 6:

(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)

There are 21 out of the entire 36 that have a sum greater than 6, hence:

P(A) = 21/36 = 7/12

The sum is divisible by 2:

(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)

There are 18 out of the entire 36 that have a sum divisible by 2, therefore:

P(B) = 18/36 = 1/2

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