Freezy's Ice Cream Stand is testing out two new flavors, Pumpernickel Brickel and Dandy Cotton Candy. A poll conducted by Freezy's showed that 32 customers liked Pumpernickel Brickel, 58 customers liked Dandy Cotton Candy, 12 liked both flavors, and 22 liked neither flavor. What is a probability that one of those customers selected at random would like Dandy Cotton Candy?

### Probability Using a Venn Diagram

It is often useful to use a Venn diagram to visualize the probabilities of multiple events. In #1 below we explore the use of a Venn diagram to determine the probabilities of individual events, the intersection of events and the compliment of an event. In #3 we will continue to explore the concept of a conditional probability and how to use a Venn diagram to solve these problems as well as the formula for conditional probability.

Let's use the Venn diagram below to find the following probabilities.

Notice that the sum of all the values in the diagram is \begin{align*}0.4 + 0.3 + 0.2 + 0.1 = 1\end{align*}. This diagram represents the entire sample space for two events, \begin{align*}A\end{align*} and \begin{align*}B\end{align*}.

- \begin{align*}P(A)\end{align*}

To find the \begin{align*}P(A)\end{align*}, we will add the probability that only \begin{align*}A\end{align*} occurs to the probability that \begin{align*}A\end{align*} and \begin{align*}B\end{align*} occur to get \begin{align*}0.4 + 0.3 = 0.7\end{align*}. So \begin{align*}P(A)=0.7\end{align*}.

- \begin{align*}P(B)\end{align*}

Similarly, \begin{align*}P(B)=0.2+0.3=0.5\end{align*}.

- \begin{align*}P(A \cap B)\end{align*}

Now, \begin{align*}P(A \cap B)\end{align*} is the value in the overlapping region 0.3.

\begin{align*}P(A \cup B)\end{align*}

\begin{align*}P(A \cup B)=0.4+0.3+0.2=0.9\end{align*}. Which can also be found using the formula \begin{align*}P(A)+P(B)-P(A \cap B)=0.7+0.5-0.3=0.9\end{align*}.

\begin{align*}P(A^\prime \cap B^\prime)\end{align*}

\begin{align*}P(A^\prime \cap B^\prime)\end{align*} needs to be determined by finding where in the diagram everything outside of \begin{align*}A\end{align*} overlaps with everything outside of \begin{align*}B\end{align*}. That will be the region outside of both circles and that probability is 0.1. Another way to think of this is \begin{align*}P(A \cup B)^\prime\end{align*}, or \begin{align*}1-P(A \cup B)\end{align*}.

#### De Morgan's Law

There are a couple of equivalent set notations on probabilities and they are called De Morgan’s Laws.

\begin{align*}(A \cap B)^\prime=(A^\prime \cup B^\prime)\end{align*} for sets or \begin{align*}P(A \cap B)^\prime=P(A^\prime \cup B^\prime)\end{align*} for probabilities.

and

\begin{align*}(A \cup B)^\prime=(A^\prime \cap B^\prime)\end{align*} for sets or \begin{align*}P(A \cup B)^\prime=P(A^\prime \cap B^\prime)\end{align*} for probabilities.

Now, let's solve the following problems.

- Given \begin{align*}P(A)=0.6\end{align*}, \begin{align*}P(B)=0.3\end{align*} and \begin{align*}P(A \cup B)=0.7\end{align*}, find \begin{align*}P(A \cap B)\end{align*} and \begin{align*}P(A^\prime \cup B^\prime)\end{align*}.

First, we can use the formula for the union of two sets to determine the intersection.

\begin{align*}P(A)+P(B)-P(A \cap B)&=P(A \cup B) \\ 0.6+0.3-P(A \cap B)&=0.7 \\ 0.9-0.7&=P(A \cap B) \\ 0.2&=P(A \cap B)\end{align*}

Now we can use De Morgan’s Law to find \begin{align*}P(A^\prime \cup B^\prime)\end{align*}.

\begin{align*}P(A^\prime \cup B^\prime)=P(A \cap B)^\prime=1-P(A \cup B)=1-0.2=0.8.\end{align*}

We could have also created a Venn diagram for the probabilities and interpreted \begin{align*}P(A^\prime \cup B^\prime)\end{align*} and the regions outside \begin{align*}A\end{align*} union with the regions outside \begin{align*}B\end{align*} which would be everything in the Venn diagram except the overlap of the two regions or \begin{align*}P(A \cap B)\end{align*}.

- The data from a survey of 140 students showed that 37 study music, 103 play a sport and 25 do neither. Create a Venn diagram to illustrate the data collected and then determine the probability that if a student is selected at random,

- he or she will study music
- he or she will study music given that he or she plays a sport.

Let \begin{align*}M\end{align*} represent the set of students who study music and \begin{align*}S\end{align*} represent the set of students who play sports. First let’s determine the number of students that study music and play a sport to fill in the overlapping region in the diagram and then we can find the other values.

\begin{align*}n(M)+n(S)-n(M \cap S)&=n(M \cup S) \\ 37+103-n(M \cap S)&=115 \\ n(M \cap S)&=25\end{align*}

- The probability that a randomly selected student studies music is the number of students who study music divided by the total number of students surveyed or \begin{align*}P(M)=\frac{n(M)}{140}=\frac{37}{140} \thickapprox 0.264\end{align*}.
- The probability that a randomly selected student will study music given that he/she plays a sport is called a conditional probability. We use the notation \begin{align*}P(M|S)\end{align*} to represent the \begin{align*}P(M)\end{align*} given that \begin{align*}S\end{align*} has already occurred. To find this probability using a Venn diagram, we find the number of student who study music and play a sport and divide by the number of students who play a sport or \begin{align*}P(M|S)=\frac{n(M \cap S)}{n(S)}=\frac{25}{103} \thickapprox 0.243\end{align*}. Think of it this way, when we say that we know that the student plays a sport, then the numerator is limited to those students who study music and play a sport and the denominator is limited to those who play a sport.

There is also a formula for conditional probability: \begin{align*}P(A|B)=\frac{P(A \cap B)}{P(B)}\end{align*}

In the context of our problem, it is: \begin{align*}P(M|S)=\frac{P(M \cap S)}{P(S)}=\frac{\frac{25}{140}}{\frac{103}{140}}=\frac{25}{140} \cdot \frac{140}{103}=\frac{25}{103}\end{align*}.

Notice that the resulting probability is the same as previously determined. Either method can be used.

### Examples

#### Example 1

Earlier, you were asked to find the probability that one of those customers selected at random would like Dandy Cotton Candy.

The Venn diagram for this situation would show 12, the number of customers who like both flavors, as the intersection. The number of customers who like only the Pumpernickel flavor would therefore be \begin{align*}32-12=20\end{align*} while the number of customers who like only the Cotton Candy flavor would be \begin{align*}58-12=46\end{align*}. Since there are 21 customers who like neither flavor, the total number of customers polled is \begin{align*}12 + 20 + 46 + 22=100\end{align*} and the probability that one of those customers chosen at random would like the Cotton Candy flavor is \begin{align*}\frac{58}{100} = 58\%\end{align*}.

#### Example 2

In a class of 260 seniors, 93 study Spanish, 95 study Chemistry, 165 study Mathematics, 18 study Spanish and Chemistry, 75 study Chemistry and Math, 20 study Math and Spanish and 15 study all three subjects. Make a Venn diagram to illustrate the data and then find the probability that a student selected at random studies:

just Spanish

\begin{align*}P(S \cap M^\prime \cap C^\prime)=\frac{70}{260}=\frac{7}{26} \thickapprox 0.269\end{align*}

Math and Chemistry but not Spanish

\begin{align*}P(M \cap C \cap S^\prime)=\frac{60}{260}=\frac{3}{13} \thickapprox 0.231\end{align*}

none of these subjects

\begin{align*}P(M^\prime \cap C^\prime \cap S^\prime)=\frac{5}{260}=\frac{1}{52} \thickapprox 0.0192\end{align*}

Spanish, given that he/she studies Math

\begin{align*}P(S|M)=\frac{P(S \cap M)}{P(M)}=\frac{\frac{20}{260}}{\frac{165}{260}}=\frac{4}{33} \thickapprox 0.121\end{align*}

#### Example 3

Given \begin{align*}P(A \cap B)=0.4\end{align*}, \begin{align*}P(A \cap B^\prime)=0.2\end{align*} and \begin{align*}P(A^\prime \cap B^\prime)=0.3\end{align*}, find \begin{align*}P(B)\end{align*} and \begin{align*}P(A|B)\end{align*}.

The information gives us the Venn diagram:

The missing value, \begin{align*}P(B \cap A^\prime)\end{align*}, must be 0.1 in order for the total of the probabilities in the sample space to equal 1. Thus, \begin{align*}P(B)=0.5\end{align*}. \begin{align*}P(A|B)=\frac{P(A \cap B)}{P(B)}=\frac{0.4}{0.5}=\frac{4}{5}=0.8\end{align*}.

### Review

For questions 1-3, find the indicated probabilities given \begin{align*}P(A)=0.5\end{align*}, \begin{align*}P(B)=0.65\end{align*} and \begin{align*}P(A \cup B)=0.75\end{align*}.

- \begin{align*}P(A \cap B)\end{align*}
- \begin{align*}P(A^\prime \cap B^\prime)\end{align*}
- \begin{align*}P(B|A)\end{align*}

For questions 4-6, find the indicated probabilities given \begin{align*}P(A)=0.6\end{align*}, \begin{align*}P(B)=0.8\end{align*} and \begin{align*}P(A \cup B)^\prime=0.2\end{align*}.

- \begin{align*}P(A \cap B^\prime)\end{align*}
- \begin{align*}P(B|A)\end{align*}
- \begin{align*}P(A|B)\end{align*}

For questions 7-9, find the indicated probabilities given \begin{align*}P(A \cap B^\prime)=0.3\end{align*}, \begin{align*}P(B \cap A^\prime)=0.2\end{align*} and \begin{align*}P(A \cup B)=0.8\end{align*}.

- \begin{align*}P(A \cap B)\end{align*}
- \begin{align*}P(A)\end{align*}
- \begin{align*}P(B|A)\end{align*}
- Given \begin{align*}P(A)=2P(B)\end{align*}, \begin{align*}P(A \cup B)=0.8\end{align*} and \begin{align*}P(A \cap B)=0.1\end{align*}, find \begin{align*}P(A)\end{align*}.
- The international club at a school has 105 members, many of whom speak multiple languages. The most commonly spoken languages in the club are English, Spanish and Chinese. Use the Venn Diagram below to determine the probability of selecting a student who:
- does not speak English.
- speaks Spanish given that he/she speaks English.
- speaks English given that he/she speaks Chinese.
- speaks Spanish and English but not Chinese.

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 12.13.