Question Video: Determining the Union of Two Mutually Exclusive Events Mathematics

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).

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Video Transcript

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

We are told in this question that our three events are mutually exclusive. We recall that two or more events are mutually exclusive if they cannot happen at the same time. This can be represented on a Venn diagram as shown, where there is no overlap or intersection between the three circles which represent the events 𝐴, 𝐵, and 𝐶. Since the sample space is equal to 𝐴 union 𝐵 union 𝐶, then the probability of 𝐴 union 𝐵 union 𝐶 is one. The sum of the probabilities inside our circles must equal one.

We also recall that if we have two mutually exclusive events 𝑋 and 𝑌, then the probability of 𝑋 union 𝑌 is equal to the probability of 𝑋 plus the probability of 𝑌. This means that in our question, the probability of 𝐴 plus the probability of 𝐵 plus the probability of 𝐶 must equal one. We are told that the probability of 𝐴 is equal to one-fifth of the probability of 𝐵. If we multiply both sides of this equation by five, we have the probability of 𝐵 is equal to five multiplied by the probability of 𝐴. We can substitute this into our equation. We can also replace the probability of 𝐶 by four multiplied by the probability of 𝐴.

Our equation becomes the probability of 𝐴 plus five multiplied by the probability of 𝐴 plus four multiplied by the probability of 𝐴 is equal to one. The left-hand side simplifies to 10 multiplied by the probability of 𝐴. We can then divide through by 10 so that the probability of 𝐴 is equal to one-tenth. This could also be written as the decimal 0.1. The probability of 𝐵 is equal to five multiplied by this. This is equal to five-tenths, which simplifies to one-half. The probability of event 𝐵 is one-half. In the same way, the probability of event 𝐶 is four-tenths. Dividing the numerator and denominator by two, this is equal to two-fifths.

We have been asked to calculate the probability of 𝐵 union 𝐶. Using our rule for mutually exclusive events, this is equal to the probability of 𝐵 plus the probability of 𝐶. Using the two fractions with a common denominator, we have five-tenths plus four-tenths, which is equal to nine-tenths. The probability of 𝐵 union 𝐶, where the two events are mutually exclusive, is nine-tenths. This is the sum of the probabilities inside the two circles in our Venn diagram.

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