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# Conditional Probability

## P(B|A) = P(A and B)/ P(A)

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Conditional Probability

### Conditional Probability

#### Notation

We know that the probability of observing an even number on a throw of a die is 0.5. Let the event of observing an even number be event \begin{align*}A\end{align*}. Now suppose that we throw the die, and we know that the result is a number that is 3 or less. Call this event \begin{align*}B\end{align*}. Would the probability of observing an even number on that particular throw still be 0.5? The answer is no, because with the introduction of event \begin{align*}B\end{align*}, we have reduced our sample space from 6 simple events to 3 simple events. In other words, since we have a number that is 3 or less, we now know that we have a 1, 2 or 3. This becomes, in effect, our sample space. Now the probability of observing a 2 is \begin{align*}\frac{1}{3}\end{align*}. With the introduction of a particular condition (event \begin{align*}B\end{align*}), we have changed the probability of a particular outcome. The Venn diagram below shows the reduced sample space for this experiment, given that event \begin{align*}B\end{align*} has occurred:

The only even number in the sample space for \begin{align*}B\end{align*} is the number 2. We conclude that the probability that \begin{align*}A\end{align*} occurs, given that \begin{align*}B\end{align*} has occurred, is 1:3, or \begin{align*}\frac{1}{3}\end{align*}. We write this with the notation \begin{align*}P(A|B)\end{align*}, which reads “the probability of \begin{align*}A\end{align*}, given \begin{align*}B\end{align*}.” So for the die toss experiment, we would write \begin{align*}P(A|B)=\frac{1}{3}\end{align*}.

#### Conditional Probability of Two Events

If \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are two events, then the probability of event \begin{align*}A\end{align*} occurring, given that event \begin{align*}B\end{align*} has occurred, is called conditional probability. We write it with the notation \begin{align*}P(A|B)\end{align*}, which reads “the probability of \begin{align*}A\end{align*}, given \begin{align*}B\end{align*}.”

To calculate the conditional probability that event \begin{align*}A\end{align*} occurs, given that event \begin{align*}B\end{align*} has occurred, take the ratio of the probability that both \begin{align*}A\end{align*} and \begin{align*}B\end{align*} occur to the probability that \begin{align*}B\end{align*} occurs. That is:

\begin{align*}P(A|B)=\frac{P(A \cap B)}{P(B)}\end{align*}

For our example above, the die toss experiment, we proceed as is shown below:

\begin{align*}& A: {\text{observe an even number}}\end{align*}

\begin{align*}& B: {\text{observe a number less than or equal to } 3}\end{align*}

To find the conditional probability, we use the formula as follows:

\begin{align*}P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{P(2)}{P(1)+P(2)+P(3)}=\frac{\frac{1}{6}}{\frac{3}{6}}=\frac{1}{3}\end{align*}

A medical research center is conducting experiments to examine the relationship between cigarette smoking and cancer in a particular city in the USA. Let \begin{align*}A\end{align*} represent an individual who smokes, and let \begin{align*}C\end{align*} represent an individual who develops cancer. This means that \begin{align*}AC\end{align*} represents an individual who smokes and develops cancer, \begin{align*}AC'\end{align*} represents an individual who smokes but does not develop cancer, and so on. We have four different possibilities, or simple events, and they are shown in the table below, along with their associated probabilities.

Simple Events Probabilities
\begin{align*}AC\end{align*} 0.10
\begin{align*}AC'\end{align*} 0.30
\begin{align*}A'C\end{align*} 0.05
\begin{align*}A'C'\end{align*} 0.55

Figure: A table of probabilities for combinations of smoking, \begin{align*}A\end{align*}, and developing cancer, \begin{align*}C\end{align*}.

These simple events can be studied, along with their associated probabilities, to examine the relationship between smoking and cancer.

#### Using Associated Probabilities to Examine the Relationship Between Two Variables

Determine the rates of developing cancer for smokers and non-smokers.

First, let's write out events symbolically:

\begin{align*}& A: {\text{individual smokes}}\end{align*}

\begin{align*}& C: {\text{individual develops cancer}}\end{align*}

\begin{align*}& A': {\text{individual does not smoke}}\end{align*}

\begin{align*}& C': {\text{individual does not develop cancer}}\end{align*}

A very powerful way of examining the relationship between cigarette smoking and cancer is to compare the conditional probability that an individual gets cancer, given that he/she smokes, with the conditional probability that an individual gets cancer, given that he/she does not smoke. In other words, we want to compare \begin{align*}P(C|A)\end{align*} with \begin{align*}P(C|A')\end{align*}.

Recall that \begin{align*}P(C|A)=\frac{P(C \cap A)}{P(A)}\end{align*}.

Before we can use this relationship, we need to calculate the value of the denominator. \begin{align*}P(A)\end{align*} is the probability of an individual being a smoker in the city under consideration. To calculate it, remember that the probability of an event is the sum of the probabilities of all its simple events. A person can smoke and have cancer, or a person can smoke and not have cancer. That is:

\begin{align*}P(A)=P(AC)+P(AC')=0.10+0.30=0.4\end{align*}

This tells us that according to this study, the probability of finding a smoker selected at random from the sample space (the city) is 40%. We can continue on with our calculations as follows:

\begin{align*}P(C|A)=\frac{P(A \cap C)}{P(A)}=\frac{P(AC)}{P(A)}=\frac{0.10}{0.40}=0.25=25\%\end{align*}

Similarly, we can calculate the conditional probability of a nonsmoker developing cancer:

\begin{align*}P(C|A')=\frac{P(A' \cap C)}{P(A')}=\frac{P(A'C)}{P(A')}=\frac{0.05}{0.60}=0.08=8\%\end{align*}

In this calculation, \begin{align*}P(A')=P(A'C)+P(A'C')=0.05+0.55=0.60.\end{align*} \begin{align*}P(A')\end{align*} can also be found by using the Complement Rule as shown: \begin{align*}P(A')=1-P(A)=1-0.40=0.60\end{align*}.

#### Determining the Likelihood of an Event

Use the calculations from the first example to determine how many more times likely a smoker is to develop cancer than a non-smoker is.

From the calculations in the first example, we can clearly see that a relationship exists between smoking and cancer. The probability that a smoker develops cancer is 25%, and the probability that a nonsmoker develops cancer is only 8%. The ratio between the two probabilities is \begin{align*}\frac{0.25}{0.08}=3.125\end{align*}, which means a smoker is more than three times more likely to develop cancer than a nonsmoker. Keep in mind, though, that it would not be accurate to say that smoking causes cancer. However, our findings do suggest a strong link between smoking and cancer.

#### Natural Frequencies Approach

There is another and interesting way to analyze this problem, which has been called the natural frequencies approach (see G. Gigerenzer, “Calculated Risks” Simon and Schuster, 2002).

#### Using the Natural Frequencies Approach

Use the natural frequencies approach to find the probability of having cancer given that you smoke.

We will use the probability information given above to demonstrate this approach. Suppose you have 1000 people. Of these 1000 people, 100 smoke and have cancer, and 300 smoke and don’t have cancer. Therefore, of the 400 people who smoke, 100 have cancer. The probability of having cancer, given that you smoke, is \begin{align*}\frac{100}{400} = 0.25\end{align*}.

Of these 1000 people, 50 don’t smoke and have cancer, and 550 don’t smoke and don’t have cancer. Thus, of the 600 people who don’t smoke, 50 have cancer. Therefore, the probability of having cancer, given that you don’t smoke, is \begin{align*}\frac{50}{600} = 0.08\end{align*}.

### Examples

In a recent election, 35% of the voters were democrats and 65% were not. Of the democrats, 75% voted for candidate Z and of the non-Democrats, 15% voted for candidate Z. Define the following events:

A = voter is Democrat

B = voted for candidate Z

#### Example 1

Find \begin{align*}P(B|A), P(B|A^c)\end{align*}

\begin{align*}P(B|A)=0.75\end{align*}

\begin{align*}P(B|A^c)=0.15\end{align*}

#### Example 2

Find \begin{align*} P(A\cap B)\end{align*} and explain in words what this represents.

\begin{align*}P(A\cap B)=P(B|A)\cdot P(A)=0.75(0.35)=0.26.\end{align*}

This is the probability of being a democrat and voting for candidate Z.

#### Example 3

Find \begin{align*} P(A^c\cap B)\end{align*} and explain in words what this represents.

\begin{align*}P(A^c\cap B)=P(B|A^c)\cdot P(A^c)=0.15(0.65)=0.0975\approx 0.10.\end{align*} This is the probability of not being a democrat and voting for candidate Z.

#### Example 4

Find \begin{align*} P(B)\end{align*}.

\begin{align*}P(B)=P(A\cap B)+P(A^c \cap B)=0.26 + 0.10=0.36.\end{align*}

### Review

For 1-5, two fair coins are tossed.

i. List all the possible outcomes in the sample space.

ii. Suppose two events are defined as follows:

\begin{align*}& A: {\text{At least one head appears}}\end{align*}

\begin{align*}& B: {\text{Only one head appears}}\end{align*}

Find the probabilities:

1. \begin{align*}P(A)\end{align*}
2. \begin{align*}P(B)\end{align*}
3. \begin{align*}P(A \cap B)\end{align*}
4. \begin{align*}P(A|B)\end{align*}
5. \begin{align*}P(B|A)\end{align*}

For 6-11, a box of six marbles contains two white marbles, two red marbles, and two blue marbles. Two marbles are randomly selected without replacement, and their colors are recorded.

i. List all the possible outcomes in the sample space.

ii. Suppose three events are defined as follows:

\begin{align*}& A: {\text{Both marbles have the same color}}\end{align*}

\begin{align*}& B: {\text{Both marbles are red}}\end{align*}

\begin{align*}& C: {\text{At least one marble is red or white}}\end{align*}

Find the probabilities:

1. \begin{align*}P(B|A)\end{align*}
2. \begin{align*}P(B|A')\end{align*}
3. \begin{align*}P(B|C)\end{align*}
4. \begin{align*}P(A|C)\end{align*}
5. \begin{align*}P(C|A')\end{align*}
6. If \begin{align*}P(A)=0.3\end{align*}, \begin{align*}P(B)=0.7\end{align*}, and \begin{align*}P(A \cap B) = 0.15\end{align*}, find \begin{align*}P(A|B)\end{align*} and \begin{align*}P(B|A)\end{align*}.
7. In a large class, 65% of the students are liberal arts majors and 35% are science majors. Twenty-five percent of the liberal arts majors are seniors while 45% of the science majors are seniors.
1. If there are 200 students in the class, how many of them are science majors?
2. If there are 200 students in the class, how many of them are science majors and seniors?
3. Create a two-way table with the row variable as the major (science or liberal arts) and the column variable as the class (senior, non-senior)
4. What percent of the class are seniors?
5. Make a tree diagram for this situation.
6. Use the tree diagram to determine the percentage of the class that is seniors.
8. In a restaurant, a waitress notices that 75% of her customers order coffee and that 30% of her customers order coffee and a croissant. What is the probability that a given customer would order a croissant, given that he/she has ordered coffee?
9. Suppose a middle school has grades 7 and 8 with the same number of students in each grade. Half of the students in grade 8 are taking algebra 1 and 25% of the students in grade 7 are taking algebra 1. Suppose that an algebra 1 student is randomly chosen to attend a math competition with a math teacher. What is the probability that the student is in grade 7?
10. If A and B are independent, determine the probability of each of the following:
1. Both A and B
2. Either A or B
3. Neither A nor B
4. A but not B
5. A given that B occurs
11. In a class has 30 students, 15 play tennis, 19 play volleyball and 2 play neither of these sports. A student is randomly selected from the class. Determine the probability that the student:
1. Plays both tennis and volleyball
2. Plays at least one of these two sports
3. Plays volleyball given that he/she does not play tennis.
12. According to the Migration Information Source and the U.S. Census Bureau, “only 35 percent of the foreign-born people in the US in 1997 were naturalized citizens, compared with 42.5% in 2007. What is the probability that two randomly selected foreign-born people in the US in 2007 were both naturalized citizens?
13. 65% of the students in a high school are female. Of the male students 11% are color-blind and of the female students 4% are color-blind. If a randomly chosen student
1. is color-blind, find the probability that the student is female.
2. Is not color-blind, find the probability that the student is male.
14. A jar contains 5 red and 4 non-red marbles. One marble is randomly drawn without replacement from the jar and its color is noted. A second marble is then drawn. What is the probability that:
1. The second marble is red?
2. The first was non-red given that the second was red?
15. Suppose that 5% of men are colorblind and 25% of women are colorblind. A person is chosen at random and that person is colorblind. What is the probability that the person is male? (Assume males and females are in equal numbers).

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### Vocabulary Language: English

TermDefinition
conditional probability The probability of a particular dependent event  given the outcome of the event on which it occurs.
conditional probability formula The conditional probability formula is P(A/B) = P(AUB)/P(B)
Dependent Events In probability situations, dependent events are events where one outcome impacts the probability of the other.
Favorable Outcome A favorable outcome is the outcome that you are looking for in an experiment.
Independent Events Two events are independent if the occurrence of one event does not impact the probability of the other event.
Multiplication Rule States that for 2 events (A and B), the probability of A and B is given by: P(A and B) = P(A) x P(B).
Mutually Exclusive Events Mutually exclusive events have no common outcomes.
Sample Space In a probability experiment, the sample space is the set of all the possible outcomes of the experiment.

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