Video Transcript
Suppose 𝐴 and 𝐵 are events in a
sample space which consists of equally likely outcomes. Given that 𝐴 contains six
outcomes, the probability of 𝐴 union 𝐵 is three-quarters, the probability of 𝐵 is
one-half, and the total number of outcomes is 20, find the probability that only one
of the events 𝐴 or 𝐵 occurs.
Let’s begin by looking at how we
can represent the probability that only one of the events 𝐴 and 𝐵 occurs on a Venn
diagram. The probability that only event 𝐴
occurs is shaded in pink. We know that this can be written
using the difference formula. It is the probability of 𝐴 minus
𝐵. And this is equal to the
probability of 𝐴 minus the probability of 𝐴 intersection 𝐵. The probability that only event 𝐵
occurs is shaded in blue. And this is denoted by the
probability of 𝐵 minus 𝐴. This is equal to the probability of
𝐵 minus the probability of 𝐴 intersection 𝐵. In order to answer this question,
we will need to find the sum of these two values.
Let’s now consider the information
we are given in this question. We are told that there are 20
outcomes in total and that event 𝐴 contains six of these. The probability of event 𝐴 is
therefore equal to six out of 20. Dividing the numerator and
denominator by two, this simplifies to three-tenths. We are also told that the
probability of 𝐴 union 𝐵 is three-quarters and the probability of 𝐵 is
one-half. We now have both the probabilities
of event 𝐴 and 𝐵 but not the probability of 𝐴 intersection 𝐵.
We can calculate this by using the
addition rule of probability, which states that the probability of 𝐴 union 𝐵 is
equal to the probability of 𝐴 plus the probability of 𝐵 minus the probability of
𝐴 intersection 𝐵. This can be rearranged as
shown. Substituting in our values, we have
the probability of 𝐴 intersection 𝐵 is equal to three-tenths plus a half minus
three-quarters. Our three fractions have a common
denominator of 20. So we can rewrite this as six over
20 plus 10 over 20 minus 15 over 20. This is equal to one over 20.
The probability of 𝐴 intersection
𝐵 is one twentieth. We can now use this value together
with the probabilities of 𝐴 and 𝐵 to calculate the probability of 𝐴 minus 𝐵 and
the probability of 𝐵 minus 𝐴. The probability of 𝐴 minus 𝐵 is
equal to three-tenths minus one twentieth. This is equal to five
twentieths. The probability of 𝐵 minus 𝐴 is
equal to one-half minus one twentieth. And this is equal to nine
twentieths. The probability that only one of
the events 𝐴 or 𝐵 occurs is therefore equal to five twentieths plus nine
twentieths. Adding our numerators gives us
fourteen twentieths, which in turn simplifies to seven-tenths or 0.7. The probability that only one of
the events 𝐴 or 𝐵 occurs is seven-tenths.
An alternative way to calculate
this would be to list the number of outcomes on our Venn diagram. Since the probability of 𝐴
intersection 𝐵 is one twentieth and there are a total of 20 outcomes, there is one
outcome in the intersection of 𝐴 and 𝐵. We were told that 𝐴 contains six
outcomes. Therefore, five of these will occur
in just event 𝐴. Since the probability of event 𝐵
is one-half, there are 10 outcomes in event 𝐵. And nine of these must occur in
just event 𝐵. As nine plus one plus five equals
15, there must be five outcomes that are not in event 𝐴 nor event 𝐵. This confirms that 14 of the
outcomes are in only one of event 𝐴 or 𝐵. And 14 out of 20 simplifies to
seven-tenths.
