MA121: Introduction to Statistics

Definition

The intersection of events $A$ and $B$, denoted $A ∩ B$, is the collection of all outcomes that are elements of both of the sets $A$ and $B$. It corresponds to combining descriptions of the two events using the word "and".

To say that the event $A ∩ B$ occurred means that on a particular trial of the experiment both $A$ and $B$ occurred. $A$ visual representation of the intersection of events $A$ and $B$ in a sample space $S$ is given in Figure 3.4 "The Intersection of Events ". The intersection corresponds to the shaded lens-shaped region that lies within both ovals.

Figure 3.4 The Intersection of Events A and B

EXAMPLE 12

In the experiment of rolling a single die, find the intersection $E \cap T$ of the events $E:$ "the number rolled is even" and $T$ : "the number rolled is greater than two".

Solution:

The sample space is $S=\{1,2,3,4,5,6\}$. Since the outcomes that are common to $E=\{2,4,6\}$ and $T=\{3,4,5,6\}$ are 4 and $6, E \cap T=\{4,6\}$.

In words the intersection is described by "the number rolled is even and is greater than two". The only numbers between one and six that are both even and greater than two are four and six, corresponding to $E \cap T$ given above.

EXAMPLE 13

A single die is rolled.

a. Suppose the die is fair. Find the probability that the number rolled is both even and greater than two.

b. Suppose the die has been "loaded" so that $P(1)=1 / 12, P(6)=3 / 12$, and the remaining four outcomes are equally likely with one another. Now find the probability that the number rolled is both even and greater than two.

Solution:

In both cases the sample space is $S=\{1,2,3,4,5,6\}$ and the event in question is the intersection $E \cap T=\{4,6\}$ of the previous example.

a. Since the die is fair, all outcomes are equally likely, so by counting we have $P(E \cap T)=2 / 6$.

b. The information on the probabilities of the six outcomes that we have so far is

$\begin{align*} \begin{array}{l|llllll} \text { Outcome } & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text { Probablity } & \frac{1}{12} & p & p & p & p & \frac{3}{12} \end{array} \end{align*}$

Since $P(1)+P(6)=4 / 12=1 / 3$ and the probabilities of all six outcomes add up to 1 ,

$\begin{align*} P(2)+P(3)+P(4)+P(5)=1-\frac{1}{3}=\frac{2}{3} \end{align*}$

Thus $4 p=2 / 3$, so $p=1 / 6$. In particular $P(4)=1 / 6$. Therefore

$\begin{align*} P(E \cap T)=P(4)+P(6)=\frac{1}{6}+\frac{3}{12}=\frac{5}{12} \end{align*}$

Definition

Events $A$ and $B$ are mutually exclusive if they have no elements in common.

For $A$ and $B$ to have no outcomes in common means precisely that it is impossible for both $A$ and $B$ to occur on a single trial of the random experiment. This gives the following rule.

Probability Rule for Mutually Exclusive Events

Events $A$ and $B$ are mutually exclusive if and only if

$P(A∩B)=0$

Any event $A$ and its complement $A^c$ are mutually exclusive, but $A$ and $B$ can be mutually exclusive without being complements.

EXAMPLE 14

In the experiment of rolling a single die, find three choices for an event $A$ so that the events $A$ and $E$ : "the number rolled is even" are mutually exclusive.

Solution:

Since $E=\{2,4,6\}$ and we want $A$ to have no elements in common with $E$, any event that does not contain any even number will do. Three choices are $\{1,3,5\}$ (the complement $E^{\complement}$, the odds), $\{1,3\}$, and $\{5\}$.