<meta http-equiv="refresh" content="1; url=/nojavascript/"> Conditional Probability | CK-12 Foundation

# 3.4: Conditional Probability

Created by: CK-12

## Learning Objectives

• Calculate the conditional probability that event $A$ occurs, given that event $B$ occurs.

Sometimes, we wish to change the probability of an event when we are bound to certain conditions. For example, we know that the probability of observing an even number on a throw of a die is $0.5$ (simple event $A$). However, suppose that we throw the die and the result is a number that is $3$ or less (simple event $B$). 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 the event $B$, we have reduced our sample space from $6$ simple events to $3$ simple events. In other words, with the introduction of a particular condition (the event $B$) we have changed the probability of a particular outcome. The Venn diagram below shows the reduced sample space for this experiment given that event $B$ has occurred.

The only even number in the sample space $B$ is the number $2$. We conclude that the probability that $A$ occurs, given that $B$ occurs is $1:3$, or $1/3$. We denote it by the symbol $P(A|B)$, which reads "the probability of $A$, given $B$". So for the die toss experiment, we write

$P(A|B) = \frac{1} {3}.$

## Conditional Probability of Two Events

Definition Conditional Probability
If $A$ and $B$ are two events, then the probability of the event $A$ to occur, given that event $B$ occurs is called a conditional probability. We denote it by the symbol $P(A|B)$, which reads "the probability of $A$ given $B$."

However, we want to show a systematic way of calculating conditional probabilities. Take the ratio of the probability of the part of $A$ that falls within the reduced sample space $B$ (i.e., the intersection of the two sample spaces $A$ and $B$) to the total probability of the reduced sample space.

To calculate the conditional probability that event $A$ occurs, given that event $B$ occurs, take the ratio of the probability that both $A$ and $B$ occur to the probability that $B$ occurs. That is,

$P(A|B) = \frac{P(A \cap B)} {P(B)}$

For our example above, the die toss experiment, we proceed as follows:

$& \text{A} = \left \{ \text{observe an even number}\right \}\\& \text{B} = \left \{\text{observe a number less than or equal to 3}\right \}$

We use the formula,

$P(A | B) = \frac{P(A \cap B)} {P(B)}$

and get,

$P(A|B) = \frac{P(A \cap B)} {P(B)} = \frac{P(2)} {P(1) + P(2) + P(3)} = \frac{1/6} {3/6} = \frac{1} {3}$

Example:

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

Simple Events Probabilities
$AC$ $0.10$
$AC'$ $0.30$
$A'C$ $0.05$
$A'C'$ $0.55$

Figure: A table of probabilities for combinations of smoking $(A)$ and developing cancer $(C)$.

How can these simple events be studied, along with their associated probabilities, to examine the relationship between smoking and cancer?

Solution:

We have

$& \text{A:} \left \{\text{individual smokes}\right \}\\& \text{C:} \left \{\text{individual develops cancer}\right \}\\& \text{A':} \left \{\text{individual does not smoke}\right \}\\& \text{C':} \left \{\text{individual does not develop cancer}\right \}$

A very powerful way of determining 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 $P(C|A)$ with $P(C|A')$:

$P(C | A) = \frac{P(A \cap C)} {P(A)}$

Before we enter our data into the formula, we need to calculate the value of the denominator. $P(A)$ is the probability of the individuals who smoke 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. Thus

$P(A) & = P(AC) + P(AC')\\& = 0.10 + 0.30 \\& = 0.40 \\& = 40 \%$

This tells us that according to this study, the probability of finding a smoker, selected at random from the sample space (the city), is $40 \%$. Continuing on with our calculations,

$P(C | A) = \frac{P(A \cap C)} {P(A)} = \frac{P(AC)} {P(A)} = \frac{0.10} {0.40} = 0.25 = 25 \%$

Similarly, we calculate the conditional probability of a nonsmoker that develops cancer:

$P(C | A') = \frac{P(A' \cap C)} {P(A')} = \frac{P(A'C)} {P(A')} = \frac{0.05} {0.60} = 0.08 = 8 \%$

Where $P(A') = P(A'C) + P(A'C') = 0.05 + 0.55 = 0.6 = 60 \%$. It is also equivalent to using the complementary relation $P(A') = 1 - P(A) = 1 - 0.40 = 0.60.$

So what is our conclusion from these calculations? We can clearly see that there exists a relationship between smoking and cancer: The probability that a smoker develops cancer is $25 \%$ and the probability that a nonsmoker develops cancer is only $8 \%$. Taking the ratio between the two probabilities, $25 \% \div 8 \% = 3.125,$ which means a smoker is more than three times more likely to develop cancer than a nonsmoker. Keep in mind, however, that it would not be accurate to say that smoking causes cancer but it does suggest a strong link between smoking and cancer.

## Lesson Summary

1. If $A$ and $B$ are two events, then the probability of the event $A$ to occur, given that event $B$ occurs is called a conditional probability. We denote it by the symbol $P(A|B),$ which reads "the probability of $A$ given $B$."
2. Conditional probability can be found with the equation $P(A|B) = \frac{P(A \cap B)} {P(B)}$.

## Review Questions

1. If $P(A) = 0.3, P(B) = 0.7,$ and $P(A \cap B) = 0.15,$ Find $P(A|B)$ and $P(B|A)$.
2. Two fair coins are tossed. i. List the possible outcomes in the sample space. ii. Two events are defined as follows: $& \text{A:} \left \{\text{At least one head appears} \right \} \\& \text{B:} \left \{\text{Only one head appears} \right \}$ Find $P(A), P(B), P(A \cap B), P(A|B),$ and $P(B|A)$
3. A box of six marbles contains two white, two red, and two blue. Two marbles are randomly selected without replacement and their colors are recorded. i. List the possible outcomes in the sample space. ii. Let the following events be defined: $& \text{A:} \left \{\text{Both marbles have the same color}\right \} \\& \text{B:} \left \{\text{Both marbles are red}\right \} \\& \text{C:} \left \{\text{At least one marble is red or white}\right \}$ Find $P(B|A), P(B|A'), P(B|C), P(A|C),$ and $P(C|A')$

1. $0.21, 0.5$
2. $3/4, 1/2, 1/2, 1, 2/3$
3. $1/3, 0, 1/14, 1/7, 1$

Feb 23, 2012

Jul 03, 2014