3.4: Conditional Probability
Learning Objective
- Calculate the conditional probability that event \begin{align*}A\end{align*}
A occurs, given that event \begin{align*}B\end{align*}B has occurred.
Introduction
In this lesson, you will learn about the concept of conditional probability and be presented with some examples of how conditional probability is used in the real world. You will also learn the appropriate notation associated with 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*}
The only even number in the sample space for \begin{align*}B\end{align*}
Conditional Probability of Two Events
If \begin{align*}A\end{align*}
To calculate the conditional probability that event \begin{align*}A\end{align*}
\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*}
Example: 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*}
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.
We have:
\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*}.
From these calculations, 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.
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).
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*}.
Lesson Summary
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*}.”
Conditional probability can be found with the equation \begin{align*}P(A|B)=\frac{P(A \cap B)}{P(B)}\end{align*}.
Another way to determine a conditional probability is to use the natural frequencies approach.
Multimedia Links
For an introduction to conditional probability (2.0), see SomaliNew, Conditonal Probability Venn Diagram (4:25).
For an explanation of how to find the probability of "And" statements and dependent events (2.0), see patrickJMT, Calculating Probability - "And" Statements, Dependent Events (5:36).
Review Questions
- 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*}.
- Two fair coins are tossed.
- List all the possible outcomes in the sample space.
- 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 \begin{align*}P(A), P(B), P(A \cap B), P(A|B),\end{align*} and \begin{align*}P(B|A)\end{align*}.
- 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.
- List all the possible outcomes in the sample space.
- 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 \begin{align*}P(B|A), P(B|A'), P(B|C), P(A|C), P(C|A'),\end{align*} and \begin{align*}P(C|A')\end{align*}.
Notes/Highlights Having trouble? Report an issue.
Color | Highlighted Text | Notes | |
---|---|---|---|
Please Sign In to create your own Highlights / Notes | |||
Show More |
Image Attributions
To add resources, you must be the owner of the section. Click Customize to make your own copy.