0.12 Probability  (Page 7/8)

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 $P\left(E\cap F\right)=2/6$

$P\left(E\mid F\right)=\frac{P\left(E\cap F\right)}{P\left(F\right)}=\frac{2/6}{3/6}=\frac{2}{3}$ .

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. $P\left(M\mid P\right)=\frac{P\left(M\cap P\right)}{P\left(P\right)}=\frac{8/\text{100}}{\text{21}/\text{100}}=\frac{8}{\text{21}}$ .

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

Since $E$ and $F$ are mutually exclusive, $P\left(E\cap F\right)=0$ . Therefore,

$P\left(E|F\right)=\frac{0}{.9}=0$ .

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 $P\left(E\right)=3/5$

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

$P\left(E\mid F\right)-\frac{3/8}{4/8}=\frac{3}{4}$ .

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