# 0.12 Probability  (Page 4/8)

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Mr. Washington is seeking a mathematics instructor's position at his favorite community college in Cupertino. His employment depends on two conditions: whether the board approves the position, and whether the hiring committee selects him. There is a 80% chance that the board will approve the position, and there is a 70% chance that the hiring committee will select him. If there is a 90% chance that at least one of the two conditions, the board approval or his selection, will be met, what is the probability that Mr. Washington will be hired?

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, $P\left(A\cap S\right)$ .

$P\left(A\cup S\right)=P\left(A\right)+P\left(S\right)-P\left(A\cap S\right)$

Substituting the known values, we get

$\text{.}\text{90}=\text{.}\text{80}+\text{.}\text{70}-P\left(A\cap S\right)$

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

The probability that this weekend will be cold is $\text{.}6$ , the probability that it will be rainy is $\text{.}7$ , and probability that it will be both cold and rainy is $\text{.}5$ . What is the probability that it will be neither cold nor rainy?

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

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

$P\left(C\cup R\right)=P\left(C\right)+P\left(R\right)-P\left(C\cap R\right)=\text{.}6+\text{.}7-\text{.}5=\text{.}8$

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

$P\left({\left(C\cup R\right)}^{c}\right)=1-P\left(C\cup R\right)=1-\text{.}8=\text{.}2$

We summarize this section by listing the important rules.

For Two Events $E$ and $F$ , $P\left(E\cup F\right)=P\left(E\right)+P\left(F\right)-P\left(E\cap F\right)$

The Addition Rule for Mutually Exclusive Events

If Two Events $E$ and $F$ are Mutually Exclusive, then $P\left(E\cup F\right)=P\left(E\right)+P\left(F\right)$

The Complement Rule

If ${E}^{c}$ is the Complement of Event $E$ , then $P\left({E}^{c}\right)=1-P\left(E\right)$

## Probability using tree diagrams and combinations

In this section, we will apply previously learnt counting techniques in calculating probabilities, and use tree diagrams to help us gain a better understanding of what is involved.

We begin with an example.

Suppose a jar contains 3 red and 4 white marbles. If two marbles are drawn with replacement, what is the probability that both marbles are red?

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 $3×3=9$ ordered pairs that show red on the first draw and, also, red on the second draw. Therefore,

$P\left(E\cap F\right)=\frac{9}{\text{49}}=\frac{3}{7}\cdot \frac{3}{7}$

Further note that in this particular case

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

If in the [link] , the two marbles are drawn without replacement, then what is the probability that both marbles are red?

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 $3×2=6$ ordered pairs that show red on the first draw and red on the second draw. Therefore,

$P\left(E\cap F\right)=\frac{6}{\text{42}}=\frac{3}{7}\cdot \frac{2}{6}$

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,

$P\left(E\cap F\right)=P\left(E\right)\cdot P\left(F\mid E\right)$

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

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