0.12 Probability  (Page 3/8)

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.

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

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.

$S=\left\{\mathrm{1,2,3,4,5,6}\right\}$ , $E=\left\{\mathrm{2,4,6}\right\}$ , $F=\left\{\mathrm{5,6}\right\}$ , 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 $n(E\cap F)$ , we remove the duplicity and get the correct answer. So as a rule,

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

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

Applying the above for this example, we get

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

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

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

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 $P\left(A\cup H\right)$ .

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

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

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

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