Conditional Probability: From Kings to Prisoners

Abstract

Again let’s roll a pair of dice once, and let’s consider the probability set-up with our now familiar 36 possible outcomes and the uniform distribution on the resulting sample space. Let A be the event “rolling a 7” in Chapter 1 we saw that this event consists of six outcomes each having probability 1/36, so P(A) = 6/36 = 1/6. Let B be the event “the first (red) die comes up 1.” If we write down all the outcomes in B, we find again six outcomes, namely, (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6). So P(B) = 1/6 too. Now suppose we compute P(A∪B). By definition A∪B is the event occurring when either 7 is rolled or a 1 appears on the first die, or both happen. Write down all the outcomes described by this situation and you will find 11 outcomes: the six ways of rolling 7, which includes the outcome (1,6), plus the other five outcomes with 1 in the first position. According to our rules for computing probabilities P(A∪B) = 11/36. How about P(A∩B)? Well, A∩B is the event occurring when both A and B happen, namely, when the first die rolls 1 and we also get 7; this can happen in one way, when (1, 6) appears. Therefore P(A∩B) = 1/36. Notice the validity of the following formula:P(A∪B) = P(A) + P(B) â?P(A∩B).