# 0.12 Probability  (Page 7/8)

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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

$P\left(E\mid F\right)=\frac{c/n}{m/n}$

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

$P\left(E\mid F\right)=\frac{P\left(E\cap F\right)}{P\left(F\right)}$

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 $P\left(E\mid F\right)$ .

$E=\left\{2,4,6\right\}$ and $F=\left\{4,5,6\right\}$ . Which implies, $E\cap F=\left\{4,6\right\}$

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.

1. $P\left(D\mid M\right)$
2. $P\left(F\mid D\right)$
3. $P\left(M\mid P\right)$

We use the conditional probability formula $P\left(E\mid F\right)=\frac{P\left(E\cap F\right)}{P\left(F\right)}$ .

1. $P\left(D\mid M\right)=\frac{P\left(D\cap M\right)}{P\left(M\right)}=\frac{\text{39}/\text{100}}{\text{47}/\text{100}}=\frac{\text{39}}{\text{47}}$ .
2. $P\left(F\mid D\right)=\frac{P\left(F\cap D\right)}{P\left(D\right)}=\frac{\text{40}/\text{100}}{\text{79}/\text{100}}=\frac{\text{40}}{\text{79}}$ .
3. $P\left(M\mid P\right)=\frac{P\left(M\cap P\right)}{P\left(P\right)}=\frac{8/\text{100}}{\text{21}/\text{100}}=\frac{8}{\text{21}}$ .

Given $P\left(E\right)=\text{.}5$ , $P\left(F\right)=.7$ , and $P\left(E\cap F\right)=.3$ . Find the following.

1. $P\left(E\mid F\right)$
2. $P\left(F\mid E\right)$ .

We use the conditional probability formula $P\left(E\mid F\right)=\frac{P\left(E\cap F\right)}{P\left(F\right)}$ .

1. $P\left(E\mid F\right)=\frac{\text{.}3}{\text{.}7}=\frac{3}{7}$ .
2. $P\left(F\mid E\right)=\text{.}3/\text{.}5=3/5$ .

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

$P\left(F\mid E\right)=\frac{P\left(E\cap F\right)}{P\left(E\right)}$

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 $F$ .

$S=\left\{\text{BBB},\text{BBG},\text{BGB},\text{BGG},\text{GBB},\text{GGB},\text{GGG}\right\}$

$E=\left\{\text{BBG},\text{BGB},\text{GBB}\right\}$ and $F=\left\{\text{BBB},\text{BBG},\text{BGB},\text{GBB}\right\}$

$E\cap F=\left\{\text{BBG},\text{BGB},\text{GBB}\right\}$

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.

1. The student uses an IBM given that he uses a Macintosh.
2. The student uses a Macintosh knowing that he uses an IBM.

Let event $I$ be that the student uses an IBM computer, and $M$ the probability that he uses a Macintosh.

1. $P\left(I\mid M\right)=\frac{\text{.}\text{20}}{\text{.}\text{50}}=\frac{2}{5}$
2. $P\left(M\mid I\right)=\frac{\text{.}\text{20}}{\text{.}\text{65}}=\frac{4}{\text{13}}$ .

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