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.
