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Study Guides > College Algebra

Computing the Probability of the Union of Two Events

We are often interested in finding the probability that one of multiple events occurs. Suppose we are playing a card game, and we will win if the next card drawn is either a heart or a king. We would be interested in finding the probability of the next card being a heart or a king. The union of two events E and F,written EFE\text{ and }F,\text{written }E\cup F, is the event that occurs if either or both events occur.

P(EF)=P(E)+P(F)P(EF)P\left(E\cup F\right)=P\left(E\right)+P\left(F\right)-P\left(E\cap F\right)
Suppose the spinner in Figure 2 is spun. We want to find the probability of spinning orange or spinning a bb.
A pie chart with six pieces with two a's colored orange, one b colored orange and another b colored red, one d colored blue, and one c colored green.Figure 2
There are a total of 6 sections, and 3 of them are orange. So the probability of spinning orange is 36=12\frac{3}{6}=\frac{1}{2}. There are a total of 6 sections, and 2 of them have a bb. So the probability of spinning a bb is 26=13\frac{2}{6}=\frac{1}{3}. If we added these two probabilities, we would be counting the sector that is both orange and a bb twice. To find the probability of spinning an orange or a bb, we need to subtract the probability that the sector is both orange and has a bb.
12+1316=23\frac{1}{2}+\frac{1}{3}-\frac{1}{6}=\frac{2}{3}
The probability of spinning orange or a bb is 23\frac{2}{3}.

A General Note: Probability of the Union of Two Events

The probability of the union of two events EE and FF (written EFE\cup F ) equals the sum of the probability of EE and the probability of FF minus the probability of EE and FF occurring together (\text{(} which is called the intersection of EE and FF and is written as EFE\cap F ).
P(EF)=P(E)+P(F)P(EF)P\left(E\cup F\right)=P\left(E\right)+P\left(F\right)-P\left(E\cap F\right)

Example 3: Computing the Probability of the Union of Two Events

A card is drawn from a standard deck. Find the probability of drawing a heart or a 7.

Solution

A standard deck contains an equal number of hearts, diamonds, clubs, and spades. So the probability of drawing a heart is 14\frac{1}{4}. There are four 7s in a standard deck, and there are a total of 52 cards. So the probability of drawing a 7 is 113\frac{1}{13}. The only card in the deck that is both a heart and a 7 is the 7 of hearts, so the probability of drawing both a heart and a 7 is 152\frac{1}{52}. Substitute P(H)=14,P(7)=113,andP(H7)=152P\left(H\right)=\frac{1}{4}, P\left(7\right)=\frac{1}{13}, \text{and} P\left(H\cap 7\right)=\frac{1}{52} into the formula.
P(E F)=P(E)+P(F)P(E F) =14+113152 =413\begin{array}{l}P\left(E{\cup }^{\text{ }}F\right)=P\left(E\right)+P\left(F\right)-P\left(E{\cap }^{\text{ }}F\right)\hfill \\ \text{ }=\frac{1}{4}+\frac{1}{13}-\frac{1}{52}\hfill \\ \text{ }=\frac{4}{13}\hfill \end{array}
The probability of drawing a heart or a 7 is 413\frac{4}{13}.

Try It 3

A card is drawn from a standard deck. Find the probability of drawing a red card or an ace. Solution

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