Question

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 succession and that the face values of the two roles are added together the sum is recorded as the outcome of a single trial of the random experiment.
Compute the probability of each of the following events:
Event A: The sum is greater than 8.
Event B: The sum is an even number.

Answer 1

Answer:  Probability of,  Event A = 5/18

Event B = 1/2

Step-by-step explanation:

Since, The probability of an event = Number of outcomes of the event/ total number of outcomes.

Since, If a fair dire rolled two times then the total number of outcomes, n(S) = 6 × 6 = 36

And, Event A: The sum is greater than 8.

So, possible outcomes are, ( 3,6), (4,5), (4,6), (5,5), (5,6), (5,4), (6,6), (6,5), (6,4), (6,3)

Thus, n(A) = 10

Since, Probability of event A,

P(A) = n(A)/n(S) = 10/36 = 5/18

Therefore, P(A)= 5/18

Now, Event B: The sum is an even number.

So, possible outcomes are, (1,1),(1,3), (1,5), (2,2), (2,4), (2,6), (3,1), (3,3), (3,5), (4,6), (4,2), (4,4), (5,1), (5,3), (5,5), (6,2), (6,4), (6,6)

Thus, n(B)= 18

Since, Probability of event B,

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

Therefore, P(B)= 1/2