How does the Addition Rule for the union of probabilities: \begin{align*} P(A \ or \ B)=P(A)+ P(B)\end{align*} work when at least some of the events overlap? For example, in a room with 20 people, there are 5 women wearing red and 5 wearing yellow, and there are 5 men wearing red and 5 wearing green. What is the probability of randomly picking one person who is either wearing red or is male?

Since \begin{align*}P(wearing \ red)\end{align*} actually overlaps with \begin{align*} P(male)\end{align*}, we can’t just use the addition rule, so how do we find the answer?

We will return to this question at the end of the lesson.

### Mutually Inclusive Events

**Mutually inclusive** events are events that have at least some amount of “overlap”, in other words, at least one of the favorable outcomes of one event is the same as a favorable outcome of another event. Because one or more outcomes may satisfy multiple cases, we cannot simply add up the probabilities as we did with

**events or we would end up with some events being counted twice. To account for the duplication, we just need to subtract the duplicated probabilities from the sum.**

*mutually exclusive*The modified formula looks like this:

\begin{align*} P(A \ or \ B)(mutually \ inclusive)&=P(A)+P(B)-P(A \ and \ B) \\ \text{Prob that either }A\text{ or }B\text{ will occur}&=\text{Prob of }A\text{ occurring}+ \text{Prob of }B\text{ occurring} - \text{Prob of both at once}\end{align*}

**Calculating Probability **

Consider a bag with five marbles in it. If there are three large marbles, one green, one blue, and one red, and also one each small red and small blue marbles, what is the probability that a random choice would be small or red?

If we were to try to solve this with the simple addition rule, we would **wrongly** get \begin{align*}\frac{2}{5}+ \frac{2}{5}=\frac{4}{5}\end{align*} *or* 80%.

In fact, if we check that answer with a table of possible outcomes, we can *see* that it is incorrect since there are only three marbles that could qualify as either small or red: 1) The large red marble, 2) The small red marble, and 3) The small blue marble.

The correct solution is:

\begin{align*}\frac{1 \ \text{large red}+2 \ \text{small}}{5 \ \text{total marbles}}=\frac{3}{5} \ or \ 60 \%\end{align*}

What went wrong when we used the simple addition rule? The problem is that we ended up counting the small red marble *twice*.

**Incorrectly** calculated: \begin{align*} \frac{ (1 \ \text{large red} +1 \ {\color{red} \text{small red}} )+ (1 \ \text{small blue}+1 \ {\color{red}\text{small red}} )}{5 \ \text{total marbles}}=\frac{4 \ marbles}{5 \ marbles}\end{align*}

**Properly** calculated: \begin{align*}\frac{ ( (1 \ \text{large red}+1 \text{ small red} )+ (1 \ \text{small blue})}{5 \ \text{total marbles}}=\frac{3 \ marbles}{5 \ marbles}\end{align*}

This leads us back to the modified addition rule for mutually inclusive events:

\begin{align*}P(A \ or \ B)=(P(A)+P(B)-P(A \ and \ B) ) \end{align*}

**Calculating Theoretical Probability **

Suppose you are playing with the spinner in the image below. What is the theoretical probability that the spinner would randomly land on either a top quadrant or a red quadrant?

There are two favorable events here, red and top. To apply the modified addition rule, we need to know the probability of each case, and the probability of the intersection of the two cases:

The probability of the spinner landing on red is:

\begin{align*}P(red)=\frac{2 \ \text{red spaces}}{4 \ \text{total spaces}}=\frac{1}{2} \ or \ 50 \%\end{align*}

The probability of the spinner landing on a top space is:

\begin{align*}P(top)=\frac{2 \ \text{top spaces}}{4 \ \text{total spaces }}=\frac{1}{2} \ or \ 50 \%\end{align*}

The probability of the spinner landing on a top red space is:

\begin{align*}P(top \ AND \ red)=\frac{1 \ \text{top red space}}{4 \ \text{total spaces}}=\frac{1}{4} \ or \ 25\%\end{align*}

Inserting those values into the formula, we get:

\begin{align*}P(top \ OR \ red)=\left(\frac{1}{2}+ \frac{1}{2} \right)-\frac{1}{4}=\frac{3}{4} \ or \ 75\%\end{align*}

**Determining Inclusivity and Exclusivity **

If \begin{align*}P(A)=40\%\end{align*}, \begin{align*}P(B)=30\%\end{align*}, and \begin{align*}P(A \ and \ B)=2\%\end{align*}, are \begin{align*} P(A)\end{align*} and \begin{align*}P(B)\end{align*} mutually inclusive or mutually exclusive?

This one is easier than it looks. If \begin{align*}P(A \ and \ B)\end{align*} is greater than 0%, then they are inclusive, since it is possible for there to be outcomes that are both \begin{align*}A\end{align*} and \begin{align*}B\end{align*}.

**Earlier Problem Revisited**

*In a room with 20 people, there are 5 women wearing red and 5 wearing yellow, and there are 5 men wearing red and 5 wearing green. What is the probability of randomly picking one person who is either wearing red or is male?*

Now we know to use the modified addition rule for inclusive events to solve this sort of problem:

\begin{align*}P(red) &=\frac{5 \ \text{women in red}+5 \ \text{men in red}} {20 \ \text{total people}}=\frac{1}{2} \ or \ 50 \% \\ P(male) &=\frac{5 \ \text{men in green} +5 \ \text{men in red}}{20 \ \text{total people}}=\frac{1}{2} \ or \ 50 \% \\ P(red,male) &=\frac{5 \ \text{men in red}}{20 \ \text{total people}}=\frac{1}{4} \ or \ 25 \% \end{align*}

Therefore:

\begin{align*}P(red \ \cup \ male)=\frac{1}{2}+\frac{1}{2}-\frac{1}{4}= \frac{3}{4} \ or \ 75 \%\end{align*}

### Examples

For examples 1 – 4, suppose you have a bag containing 5 quarters, 3 dimes, 4 nickels, 4 pennies, and 5 gold $1 coins.

#### Example 1

What is the probability that a random coin will be either silver or worth less than 10 cents?

Use the modified addition rule for inclusive events:

\begin{align*}P(A \ or \ B)&=(P(A)+P(B)- P(A \ and \ B) ) \\ P(silver \ or < 10 \ cents)&=(P(silver)+P(<10 \ cents)-P(silver \ and <10 \ cents) ) \\ &=\left(\frac{12}{21}+ \frac{8}{21}- \frac{4}{21} \right) \\ P(silver \ or < 10 \ cents)&=\frac{16}{21} \ or \ 76.2 \% \end{align*}

#### Example 2

What is the probability that a random coin will be either gold or worth more than 10 cents?

\begin{align*}P(gold \ or > 10 \ cents) &=(P(gold)+P(>10 \ cents)-P(gold \ and>10 \ cents) ) \\
&=\left(\frac{5}{21}+ \frac{10}{21}- \frac{5}{21} \right) \\
P(gold \ or > 10 \ cents)&=\frac{10}{21} \ or \ 47.6 \%
\end{align*}

#### Example 3

What is the probability that you pick two coins in a row that are each either silver or worth more than 10 cents?

This is a two-step problem, first we need to calculate the probability of a single choice being silver or worth more than 10 cents, and then we can apply the multiplication rule to calculate the total probability.

\begin{align*}P(silver \ or>10 \ cents)&=(P(silver)+P(>10 \ cents)-P(silver \ and>10 \ cents) ) \\ &=\left(\frac{12}{21}+ \frac{10}{21}- \frac{5}{21}\right) \\ P(silver \ or>10 \ cents)&=\frac{17}{21} \ or \ 81 \%\end{align*}

Now we use the multiplication rule: \begin{align*}P(A \ then \ B)=P(A) \times P(B)\end{align*}

\begin{align*} P(silver \ or > 10 \ cents \ \text{then} \ silver \ or>10 \ cents)&=0.81\times0.81=0.656 \\ P(silver \ or > 10 \ cents \ \text{then} \ silver \ or>10 \ cents)&=66 \%\end{align*}

#### Example 4

What is the probability that a random coin is either worth more than 9 cents or silver?

\begin{align*}P(>9 \ cents \ or \ silver) &=(P(>9 \ cents)+P(silver)-P(>9 \ cents \ and \ silver) ) \\
P(>9 \ cents \ or \ silver) &=\left( \frac{13}{21} + \frac{12}{21}- \frac{8}{21} \right) \\
P(>9 \ cents \ or \ silver)&=\left( \frac{17}{21}\right) \ or \ 81\%
\end{align*}

### Review

1. What is the probability that the outcome of 1 roll of a 10-sided die will be either even or greater than 5?

2. What is the probability that the outcome of one roll of a 12 sided die will be either prime or odd?

3. What is the probability of randomly pulling either a king or a heart from a standard deck?

4. What is the probability of randomly pulling either an even numbered card or a black card from a standard deck?

5. What is the probability that one roll of two standard dice will either result in a number either even or less than 7?

6. What is the probability that a randomly chosen month will either start with a “J” or have 30 days?

For problems 7 – 11, suppose you have a bag containing 4 blue, 3 green, 5 yellow, and 2 red marbles. All of the green and 2 of the yellow marbles are larger than normal, and 3 of the blue and 1 of the red marbles are smaller than normal.

7. What is the probability of randomly pulling a marble that is either large or yellow?

8. What is the probability that a randomly chosen marble is not large or is yellow?

9. What is the probability that a randomly chosen marble is either small or red?

10. What is the probability that a randomly chosen marble is either normally sized or blue?

11. What is the probability that a randomly chosen marble is small or blue?

### Review (Answers)

To view the Review answers, open this PDF file and look for section 6.5.