Question

Two fair dice are rolled.

(a) Find the conditional probability doubles are rolled, given the sum is eight.

(b) Find the conditional probability the sum is eight, given doubles are rolled.

(c) Find the probability at least one die lands on six.

(d) Find the conditional probability at least one die lands on six, given that doubles are not rolled.

Answer 1

Since two dice are rolled, there are total 36 points in the sample space viz. { (1,1), (1,2),....,(6,5), (6,6)}.

(a) Now, the outcomes favoring rolling of doubles are 6 out of 36 viz. {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)} and outcomes favoring a sum of 8 are 5 out of 36 viz. {(2,6), (3,5), (4,4), (5,3), (6,2)}.

Denoting the event of rolling a double by D and rolling a sum of 8 by E, we need to find P(D|E).

P(D|E) = P(D and E) / P(E)

Now (D and E) = {(4,4)} thus P(D and E) = 1/36

and P(E)= 5/36

Hence P(D|E)= 1/36/ 5/36 = 1/36* 36/5 = 1/5

(b) Now we need to find P(E|D)

P(E|D) = P(D and E) / P(D)

P(D) = 6/36

Thus P(E|D) = 1/36/ 6/36 = 1/36* 36/6 = 1/6

(c) Probability that at least one die lands on six = 1- P(None of the two die lands on six)

= 1- (5/6*5/6)

= 1- 25/36 = 11/36 = 0.3055

(d) Denoting the events that doubles are not rolled by D` and at least one die lands on six by S we need to find P(S|D`).

Now the outcomes favourable to event D` are 30 out of 36 since 6 outcomes favour the doubles. Hence P(D`)= 30/36= 5/6

Now the outcomes favoring event S are 11 viz. {(1,6), (2,6), (3,6), (4,6), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}. (You can notice the probability of it in (c) above given as 11/36}

Now P(S|D`)= P(S and D`)/ P(D`)

Again (S and D`) will contain all the elements of S except (6,6) since it is an outcome favoring the doubles. Hence P(S and D`) = 10/36

Thus P(S|D`) = 10/36 / 5/6 = 10/36 * 6/5 = 10/30 = 1/3