Question

Two dice are rolled. A = ‘sum of two dice equals 3’ B=‘sumoftwodiceequals7’ C=‘at

Answer 1

Presumably, the two dice are fair, so that each face has an equal 1/6 probability of being rolled. We roll 2 dice, so there are 6² = 36 possible outcomes, each with 1/36 probability of occurring.

We have

• P(A) = 2/36 = 1/18, because there are only 2 ways to roll a sum of 3 (1+2 and 2+1)

• P(B) = 6/36 = 1/6, because there are 6 ways of rolling a sum of 7 (1+6, 2+5, 3+4, 4+3, 5+2, 6+1)

• P(C) = 12/36 = 1/3, because if 1 die shows a 1, then there are 6 possible values for the other die, and vice versa)

(a) By definition of conditional probability,

P(A | C) = P(A and C) / P(C)

A and C only occur together if we had rolled either {1, 2} or {2, 1}, so ultimately

P(A and C) = P(A)

so that

P(A | C) = (1/18) / (1/3) = 3/18 = 1/6

(b) From the definition,

P(B | C) = P(B and C) / P(C)

By the same reasoning as in (a), we have to roll either {1, 6} or {6, 1} to have both B and C occur, so

P(B and C) = (2/36) / (1/3) = 6/36 = 1/6

(c) In order for two events X and Y to be independent, we need to have

P(X and Y) = P(X) P(Y)

or in terms of conditional probability,

P(X | Y) = P(X and Y) / P(Y) = P(X) P(Y) / P(Y) = P(X)

Then it follows that A and C are not independent, while B and C are independent.