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 [latex]E\text{ and }F,\text{written }E\cup F[/latex], is the event that occurs if either or both events occur.

[latex]P\left(E\cup F\right)=P\left(E\right)+P\left(F\right)-P\left(E\cap F\right)[/latex]

Suppose the spinner in Figure 2 is spun. We want to find the probability of spinning orange or spinning a [latex]b[/latex].

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 [latex]\frac{3}{6}=\frac{1}{2}[/latex]. There are a total of 6 sections, and 2 of them have a [latex]b[/latex]. So the probability of spinning a [latex]b[/latex] is [latex]\frac{2}{6}=\frac{1}{3}[/latex]. If we added these two probabilities, we would be counting the sector that is both orange and a [latex]b[/latex] twice. To find the probability of spinning an orange or a [latex]b[/latex], we need to subtract the probability that the sector is both orange and has a [latex]b[/latex].

[latex]\frac{1}{2}+\frac{1}{3}-\frac{1}{6}=\frac{2}{3}[/latex]

The probability of spinning orange or a [latex]b[/latex] is [latex]\frac{2}{3}[/latex].

A General Note: Probability of the Union of Two Events

The probability of the union of two events [latex]E[/latex] and [latex]F[/latex] (written [latex]E\cup F[/latex] ) equals the sum of the probability of [latex]E[/latex] and the probability of [latex]F[/latex] minus the probability of [latex]E[/latex] and [latex]F[/latex] occurring together [latex]\text{(}[/latex] which is called the intersection of [latex]E[/latex] and [latex]F[/latex] and is written as [latex]E\cap F[/latex] ).

[latex]P\left(E\cup F\right)=P\left(E\right)+P\left(F\right)-P\left(E\cap F\right)[/latex]

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 [latex]\frac{1}{4}[/latex]. There are four 7s in a standard deck, and there are a total of 52 cards. So the probability of drawing a 7 is [latex]\frac{1}{13}[/latex].

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 [latex]\frac{1}{52}[/latex]. Substitute [latex]P\left(H\right)=\frac{1}{4}, P\left(7\right)=\frac{1}{13}, \text{and} P\left(H\cap 7\right)=\frac{1}{52}[/latex] into the formula.

[latex]\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}[/latex]

The probability of drawing a heart or a 7 is [latex]\frac{4}{13}[/latex].

Try It 3

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

Solution