# 0.12 Probability  (Page 3/8)

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A family has three children. Determine whether the following pair of events are mutually exclusive.

$M=\left\{\text{The family has at least one boy}\right\}$
$N=\left\{\text{The family has all girls}\right\}$

Although the answer may be clear, we list both the sets.

$M=\left\{\text{BBB},\text{BBG},\text{BGB},\text{BGG},\text{GBB},\text{GBG},\text{GGB}\right\}$ and $N=\left\{\text{GGG}\right\}$

Clearly, $M\cap N=\text{Ø}$

Therefore, the events $M$ and $N$ are mutually exclusive.

We will now consider problems that involve the union of two events.

If a die is rolled, what is the probability of obtaining an even number or a number greater than four?

Let $E$ be the event that the number shown on the die is an even number, and let $F$ be the event that the number shown is greater than four.

The sample space $S=\left\{1,2,3,4,5,6\right\}$ . The event $E=\left\{2,4,6\right\}$ , and the event $F=\left\{5,6\right\}$

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

Since $P\left(E\right)=3/6$ , and $P\left(F\right)=2/6$ , a student may say $P\left(E\cup F\right)=3/6+2/6$ . This will be incorrect because the element 6, which is in both $E$ and $F$ has been counted twice, once as an element of $E$ and once as an element of $F$ . In other words, the set $E\cup F$ has only four elements and not five. Therefore, $P\left(E\cup F\right)=4/6$ and not $5/6$ .

This can be illustrated by a Venn diagram.

The sample space $S$ , the events $E$ and $F$ , and $E\cap F$ are listed below.

$S=\left\{1,2,3,4,5,6\right\}$ , $E=\left\{2,4,6\right\}$ , $F=\left\{5,6\right\}$ , and $E\cap F=\left\{6\right\}$ .

The above figure shows $S$ , $E$ , $F$ , and $E\cap F$ .

Finding the probability of $E\cup F$ , is the same as finding the probability that $E$ will happen, or $F$ will happen, or both will happen. If we count the number of elements $n\left(E\right)$ in $E$ , and add to it the number of elements $n\left(F\right)$ in $F$ , the points in both $E$ and $F$ are counted twice, once as elements of $E$ and once as elements of $F$ . Now if we subtract from the sum, $n\left(E\right)+n\left(F\right)$ , the number $n\left(E\cap F\right)$ , we remove the duplicity and get the correct answer. So as a rule,

$n\left(E\cup F\right)=n\left(E\right)+n\left(F\right)–n\left(E\cap F\right)$

By dividing the entire equation by $n\left(S\right)$ , we get

$\frac{n\left(E\cup F\right)}{n\left(S\right)}=\frac{n\left(E\right)}{n\left(S\right)}+\frac{n\left(F\right)}{n\left(S\right)}–\frac{n\left(E\cap F\right)}{n\left(S\right)}$

Since the probability of an event is the number of elements in that event divided by the number of all possible outcomes, we have

$P\left(E\cup F\right)=P\left(E\right)+P\left(F\right)–P\left(E\cap F\right)$

Applying the above for this example, we get

$P\left(E\cup F\right)=3/6+2/6-1/6=4/6$

This is because, when we add $P\left(E\right)$ and $P\left(F\right)$ , we have added $P\left(E\cap F\right)$ twice. Therefore, we must subtract $P\left(E\cap F\right)$ , once.

This gives us the general formula, called the Addition Rule , for finding the probability of the union of two events. It states

$P\left(E\cup F\right)=P\left(E\right)+P\left(F\right)–P\left(E\cap F\right)$

If two events E and F are mutually exclusive, then $E\cap F=\varnothing$ and $P\left(E\cap F\right)=0$ , and we get

$P\left(E\cup F\right)=P\left(E\right)+P\left(F\right)$

If a card is drawn from a deck, use the addition rule to find the probability of obtaining an ace or a heart.

Let $A$ be the event that the card is an ace, and $H$ the event that it is a heart.

Since there are four aces, and thirteen hearts in the deck, $P\left(A\right)=4/\text{52}$ and $P\left(H\right)=\text{13}/\text{52}$ . Furthermore, since the intersection of two events is an ace of hearts, $P\left(A\cap H\right)=1/\text{52}$

We need to find $P\left(A\cup H\right)$ .

$P\left(A\cup H\right)=P\left(A\right)+P\left(H\right)–P\left(A\cap H\right)=4/\text{52}+\text{13}/\text{52}-1/\text{52}=\text{16}/\text{52}$ .

Two dice are rolled, and the events $F$ and $T$ are as follows:

$F=\left\{\text{The sum of the dice is four}\right\}$ and $T=\left\{\text{At least one die shows a three}\right\}$

Find $P\left(F\cup T\right)$ .

We list $F$ and $T$ , and $F\cap T$ as follows:

$F=\left\{\left(1,3\right),\left(2,2\right),\left(3,1\right)\right\}$
$T=\left\{\left(3,1\right),\left(3,2\right),\left(3,3\right),\left(3,4\right),\left(3,5\right),\left(3,6\right),\left(1,3\right),\left(2,3\right),\left(4,3\right),\left(5,3\right),\left(6,3\right)\right\}$
$F\cap T=\left\{\left(1,3\right),\left(3,1\right)\right\}$

Since $P\left(F\cup T\right)=P\left(F\right)+P\left(T\right)-P\left(F\cap T\right)$

We have $P\left(F\cup T\right)=3/\text{36}+\text{11}/\text{36}-2/\text{36}=\text{12}/\text{36}$ .

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