Question Video: Determining the Union of Two Mutually Exclusive Events

Suppose 𝐴, 𝐵, and 𝐶 are three mutually exclusive events in a sample space 𝑆. Given that 𝑆 = 𝐴 ⋃ 𝐵 ⋃ 𝐶, 𝑃(𝐴) = (1/5)𝑃(𝐵), and 𝑃(𝐶) = 4𝑃(𝐴), find 𝑃(𝐵 ⋃ 𝐶).

02:15

Video Transcript

Suppose 𝐴, 𝐵, and 𝐶 are three mutually exclusive events in a sample space 𝑆. Given that 𝑆 is the union of 𝐴, 𝐵, and 𝐶, the probability of 𝐴 is one-fifth the probability of 𝐵, and the probability of 𝐶 is four times the probability of 𝐴, find the probability of 𝐵 union 𝐶.

The first key piece of information we’re given is that the sample space is the union of these three mutually exclusive events. This means that these three events entirely partition the sample space with no overlap. And so, the sum of their probabilities is one. We’re also told that the probability of 𝐴 is one-fifth the probability of 𝐵 and the probability of 𝐶 is four times the probability of 𝐴, so it’s four-fifths times the probability of 𝐵. We can therefore form an equation using only the probability of 𝐵. One-fifth the probability of 𝐵 plus the probability of 𝐵 plus four-fifths the probability of 𝐵 is equal to one.

If we simplify, we have one-fifth, one, and four-fifths as coefficients. So, we have that twice the probability of 𝐵 is equal to one. And dividing through by two, we find that the probability of 𝐵 is one-half. Now, the question asked us for the probability of 𝐵 union 𝐶, and so we need to recall the addition rule for mutually exclusive events. This tells us that the probability of their union is equal to the sum of their individual probabilities. We know the probability of 𝐵. It’s one-half, so we just need to calculate the probability of 𝐶.

We’ve already expressed the probability of 𝐶 as four-fifths the probability of 𝐵, so it’s four-fifths multiplied by one-half, which is four-tenths. Using the addition rule for mutually exclusive events then, the probability of 𝐵 union 𝐶 is one-half plus four-tenths. And thinking of one-half as the equivalent fraction five-tenths, we have a total probability of nine-tenths. Remember, we could only apply the addition rule and indeed the fact that the sum of the three probabilities was one because the three events 𝐴, 𝐵, and 𝐶 were mutually exclusive.