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The example we just considered consisted of only one outcome of the sample space. We are often interested in finding probabilities of several outcomes represented by an event.
An event is a subset of a sample space. If an event consists of only one outcome, it is called a simple event .
If two dice are rolled, find the probability that the sum of the faces of the dice is 7.
Let $E$ represent the event that the sum of the faces of two dice is 7.
Since the possible cases for the sum to be 7 are: (1, 6), (2,5), (3, 4), (4, 3), (5, 2), and (6, 1).
and the probability of the event $E$ ,
$P\left(E\right)=6/\text{36}$ or $1/6$ .
A jar contains 3 red, 4 white, and 3 blue marbles. If a marble is chosen at random, what is the probability that the marble is a red marble or a blue marble?
We assume the marbles are ${r}_{1}$ , ${r}_{2}$ , ${r}_{3}$ , ${w}_{1}$ , ${w}_{2}$ , ${w}_{3}$ , ${w}_{4}$ , ${b}_{1}$ , ${b}_{2}$ , ${b}_{3}$ . Let the event $C$ represent that the marble is red or blue.
The sample space $S=\left\{{r}_{1},{r}_{2},{r}_{\mathrm{3,}}{w}_{1},{w}_{2},{w}_{3},{w}_{4},{b}_{1},{b}_{2},{b}_{3}\right\}$
And the event $C=\left\{{r}_{1},{r}_{2},{r}_{3},{b}_{1},{b}_{2},{b}_{3}\right\}$
Therefore, the probability of $C$ ,
$P\left(C\right)=6/\text{10}$ or $3/5$ .
A jar contains three marbles numbered 1, 2, and 3. If two marbles are drawn, what is the probability that the sum of the numbers is 4?
Since two marbles are drawn, the sample space consists of the following six possibilities.
Let the event F represent that the sum of the numbers is four. Then
Therefore, the probability of $F$ is
$P\left(F\right)=2/6$ or $1/3$ .
A jar contains three marbles numbered 1, 2, and 3. If two marbles are drawn, what is the probability that the sum of the numbers is at least 4?
The sample space, as in [link] , consists of the following six possibilities.
Let the event $A$ represent that the sum of the numbers is at least four. Then
Therefore, the probability of $F$ is
$P\left(F\right)=4/6$ or $2/3$ .
In the [link] , we learned to find the union, intersection, and complement of a set. We will now use these set operations to describe events.
The union of two events $E$ and $F$ , $E\cup F$ , is the set of outcomes that are in $E$ or in $F$ or in both.
The intersection of two events $E$ and $F$ , $E\cap F$ , is the set of outcomes that are in both $E$ and $F$ .
The complement of an event $E$ , denoted by ${E}^{c}$ , is the set of outcomes in the sample space $S$ that are not in $E$ . It is worth noting that $P\left({E}^{C}\right)=1-P\left(E\right)$ . This follows from the fact that if the sample space has $n$ elements and $E$ has $k$ elements, then ${E}^{c}$ has $n-k$ elements. Therefore,
$P\left({E}^{C}\right)=\frac{n-k}{n}=1-\frac{k}{n}=1-P\left(E\right)$ .
Of particular interest to us are the events whose outcomes do not overlap. We call these events mutually exclusive.
Two events $E$ and $F$ are said to be mutually exclusive if they do not intersect. That is, $E\cap F=\mathrm{\varnothing}$ .
Next we'll determine whether a given pair of events are mutually exclusive.
A card is drawn from a standard deck. Determine whether the pair of events given below is mutually exclusive.
Clearly the ace of hearts belongs to both sets. That is
$E\cap F=\left\{\text{Ace of hearts}\right\}\ne \mathrm{\varnothing}$ .
Therefore, the events $E$ and $F$ are not mutually exclusive.
Two dice are rolled. Determine whether the pair of events given below is mutually exclusive.
For clarity, we list the elements of both sets.
Clearly, $G\cap H=\left\{\left(\mathrm{2,4}\right),\left(\mathrm{4,2}\right)\right\}\ne \text{\xd8}$ .
Therefore, the two sets are not mutually exclusive.
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