0.12 Probability  (Page 8/8)

1. Clearly, $P\left(\text{The card is a king}\right)=4/\text{52}=1/\text{13}$ .

2. To find $P\left(\text{The card is a king}\mid \text{A red card has shown}\right)$ , we reason as follows:

   Since a red card has shown, there are only twenty six possibilities. Of the 26 red cards, there are two kings. Therefore,

   $P\left(\text{The card is a king}\mid \text{A red card has shown}\right)=2/\text{26}=1/\text{13}$ .

According to the test, $C$ and $M$ are independent if and only if $P\left(C\cap M\right)=P\left(C\right)P\left(M\right)$ .

and $P\left(C\cap M\right)=\text{.}\text{06}$

Clearly

Therefore, the two events are not independent. We may say they are dependent.

Let $M$ be the event that a woman wears makeup, and $L$ the event that a woman has a low self-image. We have

$P\left(M\cap L\right)=9/\text{100}$ , $P\left(M\right)=\text{45}/\text{100}$ and $P\left(L\right)=\text{20}/\text{100}$

In order for two events to be independent, we must have

Since $9/\text{100}=\left(\text{45}/\text{100}\right)\left(\text{20}/\text{100}\right)$

The two events "wearing makeup" and "having a low self-image" are independent.

To make things easier, we list the sample space, the events, their intersections and the corresponding probabilities.

$E=\left\{\text{HHH},\text{HHT},\text{HTH},\text{HTT}\right\}$ , $P\left(E\right)=4/8$ or $1/2$

$F=\left\{\text{HHH},\text{HHT},\text{HTH},\text{THH}\right\}$ , $P\left(F\right)=4/8$ or $1/2$

$G=\left\{\text{HHT},\text{THH}\right\}$ , $P\left(G\right)=2/8$ or $1/4$

$E\cap F=\left\{\text{HHH},\text{HHT},\text{HTH}\right\}$ , $P\left(E\cap F\right)=3/8$

$E\cap G=\left\{\text{HHT},\text{THH}\right\}$ , $P\left(F\cap G\right)=2/8$ or $1/4$

$E\cap G=\left\{\text{HHT}\right\}$ $P\left(E\cap G\right)=1/8$

1. In order for $E$ and $F$ to be independent, we must have

   $P\left(E\cap F\right)=P\left(E\right)P\left(F\right)$ .

   But

   Therefore, $E$ and $F$ are not independent.

2. $F$ and $G$ will be independent if

   $P\left(F\cap G\right)=P\left(F\right)P\left(G\right)$ .

   Since $1/4\ne 1/2\cdot 1/4$

   $F$ and $G$ are not independent.

3. We look at

   $P\left(E\cap G\right)=P\left(E\right)P\left(G\right)$
   $1/8=1/2\cdot 1/4$

   Therefore, $E$ and $G$ are independent events.

Let $A$ be the event that Jaime will visit his aunt this year, and $R$ be the event that he will go river rafting.

We are given $P\left(A\right)=\text{.}\text{30}$ and $P\left(R\right)=\text{.}\text{50}$ , and we want to find $P\left(A\cap R\right)$ .

Since we are told that the events $A$ and $R$ are independent,

If $A$ and $B$ are independent, then by definition $P\left(B\mid A\right)=P\left(B\right)$

Therefore, $P\left(B\right)=\text{.}4$

By definition $P\left(B\mid A\right)=\frac{P\left(A\cap B\right)}{P\left(A\right)}$

Substituting, we have

Therefore, $P\left(A\cap B\right)=\text{.}\text{35}$

The addition rule states that

Since $A$ and $B$ are independent, $P(A\cap B)=P\left(A\right)P\left(B\right)$

We substitute for $P(A\cap B)$ in the addition formula and get

By letting $P\left(B\right)=x$ , and substituting values, we get

Therefore, $P\left(B\right)=.4$ .