0.12 Probability  (Page 4/8)

Let $A$ be the event that the board approves the position, and S be the event that Mr. Washington gets selected. We have,

$P\left(A\right)=\text{.}\text{80}$ , $P\left(S\right)=\text{.}\text{70}$ , and $P\left(A\cup S\right)=\text{.}\text{90}$ .

We need to find, $P\left(A\cap S\right)$ .

The addition formula states that,

Substituting the known values, we get

Therefore, $P\left(A\cap S\right)=\text{.}\text{60}$ .

Let $C$ be the event that the weekend will be cold, and $R$ be event that it will be rainy. We are given that

$P\left(C\right)=\text{.}6$ , $P\left(R\right)=\text{.}7$ , $P\left(C\cap R\right)=\text{.}5$

We want to find $P\left({\left(C\cup R\right)}^c\right)$ .

Let $E$ be the event that the first marble drawn is red, and let $F$ be the event that the second marble drawn is red.

We need to find $P\left(E\cap F\right)$ .

By the statement, "two marbles are drawn with replacement," we mean that the first marble is replaced before the second marble is drawn.

There are 7 choices for the first draw. And since the first marble is replaced before the second is drawn, there are, again, seven choices for the second draw. Using the multiplication axiom, we conclude that the sample space $S$ consists of 49 ordered pairs. Of the 49 ordered pairs, there are $3\times 3=9$ ordered pairs that show red on the first draw and, also, red on the second draw. Therefore,

Further note that in this particular case

By the statement, "two marbles are drawn without replacement," we mean that the first marble is not replaced before the second marble is drawn.

Again, we need to find $P\left(E\cap F\right)$ .

There are, again, 7 choices for the first draw. And since the first marble is not replaced before the second is drawn, there are only six choices for the second draw. Using the multiplication axiom, we conclude that the sample space $S$ consists of 42 ordered pairs. Of the 42 ordered pairs, there are $3\times 2=6$ ordered pairs that show red on the first draw and red on the second draw. Therefore,

Here $3/7$ represents $P\left(E\right)$ , and $2/6$ represents the probability of drawing a red on the second draw, given that the first draw resulted in a red. We write the latter as $P\left(\text{Red on the second}\mid \text{red on first}\right)$ or $P\left(F\mid E\right)$ . The "|" represents the word "given." Therefore,

The above result is an important one and will appear again in later sections.