Question

Q2. Two fair dice are tossed and recorded

(a) What is the probability that the sum of the two dice is at most 10?

(b) Given that the sum is an even number, what is the probability that the sum of two dice is 6 or 10?

Answer 1

When two fair dice are rolled, sample space S will be as follows : [All possible options on the face of the dice (D1,D2)]

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)}

Number of sample size n(S) = 36

a. The probability that the sum of the two dice is at most 10.

Let the event that the sum of the two dice is at the most 10 be A

Probability of A= P(A)= No of favourable events / Total number of events in Sample space

A = { (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), (6,1,), (6,2), (6,3), (6,4)}

n(A) = 33 [ the only three combinations where the sum is greater than 10 are (5,6), (6,5), (6,6)

P(A)= n(A) / n(S)

P(A)= 33/36

P(A)= 11/12 = 0.9167

b. Given that the sum is even number what is the probability that the sum of two dice is 6 or 10.

Let the event that the sum is event be B

B= { (1,1,), (1,3), (1,5), (2,2), (2,4), (2,6), (3,1,), (3,3), (3,5), (4,2), (4,4), (4,6), (5,1,), (5,3), (5,5), (6,2), (6,4), (6,6)}

P(B) = n(B) / n(S)

P(B) = 18/36 = 1/2 = 0.5

Let C be the event that the sum of two dice is 6 or 10.

C= { (1,5), (2,4), (3,3), (4,2), (4,6), (5,1,),(5,5), (6,4)}

P(C) = n(C) /n(S) = 8/36 = 2/ 9 = 0.2222

Probability that sum is 6 0r 10, given that the sum is even is conditional probability. We need to find probability of event C given that event B has occurred.

  

So the probability of getting a sum 6 or 10 when it is given that the sum is even is 4/9