Question Video: Determining the Probability of the Union of Two Events Mathematics

Suppose 𝐴 and 𝐵 are events in a sample space which consists of equally likely outcomes. Given that 𝐴 contains 6 outcomes, 𝑃(𝐴 ∪ 𝐵) = 3/4, 𝑃(𝐵) = 1/2, and the total number of outcomes is 20, find the probability that only one of the events 𝐴 or 𝐵 occurs.

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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.

We will begin by sketching a Venn diagram to model the scenario. We are trying to find the probability that only one of events 𝐴 or 𝐵 occurs. This means that we want event 𝐴 to occur or event 𝐵 to occur but not both events. The probability that only event 𝐴 occurs is written the probability of 𝐴 minus 𝐵, as this is the probability that 𝐴 occurs but 𝐵 does not. The difference rule of probability tells us that this is equal to the probability of 𝐴 minus the probability of 𝐴 intersection 𝐵. We can visualize this on the Venn diagram. It is the whole of circle 𝐴 minus the part of the circle that intersects circle 𝐵.

Likewise, the probability that only event 𝐵 occurs, written probability of 𝐵 minus 𝐴, is equal to the probability of 𝐵 minus the probability of 𝐴 intersection 𝐵. If we know the probability of 𝐴, the probability of 𝐵, and the probability of 𝐴 intersection 𝐵, then we can calculate the probability that only one of the two events occurs. We will need to add the probability of 𝐴 minus 𝐵 and the probability of 𝐵 minus 𝐴.

We are told in the question that 𝐴 contains six outcomes and that there are 20 outcomes in total. This means that the probability of event 𝐴 is six out of 20, which simplifies to three out of 10 or three-tenths. We are also told that the probability of 𝐵 is one-half. However, we are not told in the question the probability of 𝐴 intersection 𝐵. We are told the probability of 𝐴 union 𝐵. And the addition rule of probability 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 so that the probability of 𝐴 intersection 𝐵 is equal to the probability of 𝐴 plus the probability of 𝐵 minus the probability of 𝐴 union 𝐵. Substituting in our values, this is equal to three-tenths plus one-half minus three-quarters.

In order to add and subtract fractions, we need a common denominator. In this case, the lowest common multiple of 10, two, and four is 20. So we have six twentieths plus ten twentieths minus fifteen twentieths. This is equal to one twentieth. The probability of 𝐴 intersection 𝐵 is one twentieth. In the context of this question, this means that one out of the 20 outcomes occur in event 𝐴 and event 𝐵.

We now have all three values required to answer the question. The probability of 𝐴 minus 𝐵 is equal to three-tenths minus one twentieth. This is the same as six twentieths minus one twentieth, which is equal to five twentieths. Whilst this simplifies to one-quarter, we will leave the denominator as 20 at present. In the context of our question, there are five outcomes that occur in just event 𝐴. This makes sense as this gives us a total of six outcomes in event 𝐴.

The probability of 𝐵 minus 𝐴 is equal to one-half minus one twentieth. As one-half is the same as ten twentieths, the probability of 𝐵 minus 𝐴 is nine twentieths. Nine outcomes occur in just event 𝐵.

The probability that only one of the events 𝐴 or 𝐵 occurs is therefore equal to five twentieths plus nine twentieths. This is equal to fourteen twentieths. This ties in with our Venn diagram, which shows that 14 out of the 20 outcomes occur in only one of event 𝐴 or event 𝐵. The fraction fourteen twentieths can be simplified to seven-tenths or written as a decimal 0.7. This is the probability that only one of the events 𝐴 or 𝐵 occurs.

For completeness, we should add a five outside of the two circles in our Venn diagram. This is because there are a total of 20 outcomes in the sample space. Five plus one plus nine equals 15, and subtracting this from 20 gives us five. There are five outcomes that are not in event 𝐴 or event 𝐵.

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