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Let us now develop a formula for the conditional probability $P\left(E\mid F\right)$ .
Suppose an experiment consists of $n$ equally likely events. Further suppose that there are $m$ elements in $F$ , and $c$ elements in $E\cap F$ , as shown in the following Venn diagram.
If the event $F$ has occurred, the set of all possible outcomes is no longer the entire sample space, but instead, the subset $F$ . Therefore, we only look at the set $F$ and at nothing outside of $F$ . Since $F$ has $m$ elements, the denominator in the calculation of $P\left(E\mid F\right)$ is m. We may think that the numerator for our conditional probability is the number of elements in $E$ . But clearly we cannot consider the elements of $E$ that are not in $F$ . We can only count the elements of $E$ that are in $F$ , that is, the elements in $E\cap F$ . Therefore,
$P\left(E\mid F\right)=\frac{c}{m}$
Dividing both the numerator and the denominator by $n$ , we get
But $c/n=P\left(E\cap F\right)$ , and $m/n=P\left(F\right)$ .
Substituting, we derive the following formula for $P\left(E\mid F\right)$ .
For Two Events $E$ and $F$ , the Probability of $E$ Given $F$ is
A single die is rolled. Use the above formula to find the conditional probability of obtaining an even number given that a number greater than three has shown.
Let $E$ be the event that an even number shows, and $F$ be the event that a number greater than three shows. We want $P\left(E\mid F\right)$ .
$E=\left\{\mathrm{2,4,6}\right\}$ and $F=\left\{\mathrm{4,5,6}\right\}$ . Which implies, $E\cap F=\left\{\mathrm{4,6}\right\}$
Therefore, $P\left(F\right)=3/6$ , and $P\left(E\cap F\right)=2/6$
$P\left(E\mid F\right)=\frac{P\left(E\cap F\right)}{P\left(F\right)}=\frac{2/6}{3/6}=\frac{2}{3}$ .
The following table shows the distribution by gender of students at a community college who take public transportation and the ones who drive to school.
Male(M) | Female(F) | Total | |
Public Transportation(P) | 8 | 13 | 21 |
Drive(D) | 39 | 40 | 79 |
Total | 47 | 53 | 100 |
The events $M$ , $F$ , $P$ , and $D$ are self explanatory. Find the following probabilities.
We use the conditional probability formula $P\left(E\mid F\right)=\frac{P\left(E\cap F\right)}{P\left(F\right)}$ .
Given $P\left(E\right)=\text{.}5$ , $P\left(F\right)=.7$ , and $P\left(E\cap F\right)=.3$ . Find the following.
We use the conditional probability formula $P\left(E\mid F\right)=\frac{P\left(E\cap F\right)}{P\left(F\right)}$ .
Given two mutually exclusive events $E$ and $F$ such that $P\left(E\right)=\text{.}4$ , $P\left(F\right)=\text{.}9$ . Find $P\left(E\mid F\right)$ .
Since $E$ and $F$ are mutually exclusive, $P\left(E\cap F\right)=0$ . Therefore,
$P\left(E|F\right)=\frac{0}{.9}=0$ .
Given $P\left(F\mid E\right)=\text{.}5$ , and $P\left(E\cap F\right)=\text{.}3$ . Find $P\left(E\right)$ .
Using the conditional probability formula $P\left(E\mid F\right)=\frac{P\left(E\cap F\right)}{P\left(F\right)}$ , we get
Substituting,
$\text{.}5=\frac{\text{.}3}{P\left(E\right)}$ or $P\left(E\right)=3/5$
In a family of three children, find the conditional probability of having two boys and a girl, given that the family has at least two boys.
Let event $E$ be that the family has two boys and a girl, and let $F$ be the probability that the family has at least two boys. We want $P\left(E\mid F\right)$ .
We list the sample space along with the events $E$ and $F$ .
$E=\left\{\text{BBG},\text{BGB},\text{GBB}\right\}$ and $F=\left\{\text{BBB},\text{BBG},\text{BGB},\text{GBB}\right\}$
Therefore, $P\left(F\right)=4/8$ , and $P\left(E\cap F\right)=3/8$ .
And
$P\left(E\mid F\right)-\frac{3/8}{4/8}=\frac{3}{4}$ .
At a community college 65% of the students use IBM computers, 50% use Macintosh computers, and 20% use both. If a student is chosen at random, find the following probabilities.
Let event $I$ be that the student uses an IBM computer, and $M$ the probability that he uses a Macintosh.
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