MA121: Introduction to Statistics

Definition

The union of events $A$ and $B$, denoted $A \cup B$, is the collection of all outcomes that are elements of one or the other of the sets $A$ and $B$, or of both of them. It corresponds to combining descriptions of the two events using the word "or".

To say that the event $A \cup B$ occurred means that on a particular trial of the experiment either $A$ or $B$ occurred (or both did). A visual representation of the union of events $A$ and $B$ in a sample space $S$ is given in Figure $3.5$ "The Union of Events". The union corresponds to the shaded region.

Figure 3.5 The Union of Events $A$ and $B$

EXAMPLE 15

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

Solution:

Since the outcomes that are in either $E=\{2,4,6\}$ or $T=\{3,4,5,6\}$ (or both) are $2,3,4,5$, and 6 , $E \cup T=\{2,3,4,5,6\}$. Note that an outcome such as 4 that is in both sets is still listed only once (although strictly speaking it is not incorrect to list it twice).

In words the union is described by "the number rolled is even or is greater than two". Every number between one and six except the number one is either even or is greater than two, corresponding to $E \cup T$ given above.

EXAMPLE 16

A two-child family is selected at random. Let $B$ denote the event that at least one child is a boy, let $D$ denote the event that the genders of the two children differ, and let $M$ denote the event that the genders of the two children match. Find $B \cup D$ and $B \cup M$.

Solution:

A sample space for this experiment is $S=\{b b, b g, g b, g g\}$, where the first letter denotes the gender of the firstborn child and the second letter denotes the gender of the second child. The events $B, D$, and $M$ are

$\begin{align*} B=\{b b, b g, g b\} \quad D=\{b g, g b\} \quad M=\{b b, g g\} \end{align*}$

Each outcome in $D$ is already in $B$, so the outcomes that are in at least one or the other of the sets $B$ and $D$ is just the set $B$ itself: $B \cup D=\{b b, b g, g b\}=B$.

Every outcome in the whole sample space $S$ is in at least one or the other of the sets $B$ and $M$, so $B \cup M=\{b b, b g, g b, g g\}=S$

The following Additive Rule of Probability is a useful formula for calculating the probability of $A \cup B$

Additive Rule of Probability

$\begin{align*} P(A \cup B)=P(A)+P(B)-P(A \cap B) \end{align*}$

The next example, in which we compute the probability of a union both by counting and by using the formula, shows why the last term in the formula is needed.

EXAMPLE 17

Two fair dice are thrown. Find the probabilities of the following events:

Solution:

As was the case with tossing two identical coins, actual experience dictates that for the sample space to have equally likely outcomes we should list outcomes as if we could distinguish the two dice. We could imagine that one of them is red and the other is green. Then any outcome can be labeled as a pair of numbers as in the following display, where the first number in the pair is the number of dots on the top face of the green die and the second number in the pair is the number of dots on the top face of the red die.

a. There are 36 equally likely outcomes, of which exactly one corresponds to two fours, so the probability of a pair of fours is 1/36.

b. From the table we can see that there are 11 pairs that correspond to the event in question: the six pairs in the fourth row (the green die shows a four) plus the additional five pairs other than the pair 44, already counted, in the fourth column (the red die is four), so the answer is $11 / 36$. To see how the formula gives the same number, let $A_{G}$ denote the event that the green die is a four and let $A_{R}$ denote the event that the red die is a four. Then clearly by counting we get $P\left(A_{G}\right)=6 / 36$ and $P\left(A_{R}\right)=6 / 36 .$ Since $A_{G} \cap A_{R}=\{44\}, P\left(A_{G} \cap A_{R}\right)=1 / 36 ;$ this is the computation in part (a), of course. Thus by the Additive Rule of Probability,

$\begin{align*} P\left(A_{G} \cup A_{R}\right)=P\left(A_{G}\right)+P\left(A_{R}\right)-P\left(A_{G}-A_{R}\right)=\frac{6}{36}+\frac{6}{36}-\frac{1}{36}=\frac{11}{36} \end{align*}$

EXAMPLE 18

A tutoring service specializes in preparing adults for high school equivalence tests. Among all the students seeking help from the service, 63% need help in mathematics, 34% need help in English, and 27% need help in both mathematics and English. What is the percentage of students who need help in either mathematics or English?

Solution:

Imagine selecting a student at random, that is, in such a way that every student has the same chance of being selected. Let $M$ denote the event "the student needs help in mathematics" and let $E$ denote the event "the student needs help in English". The information given is that $P(M)=0.63$, $P(E)=0.34$, and $P(M \cap E)=0.27$. The Additive Rule of Probability gives

$\begin{align*} P(M \cup E)=P(M)+P(E)-P(M \cap E)=0.63+0.34-0.27=0.70 \end{align*}$

Note how the naive reasoning that if $63 \%$ need help in mathematics and $34 \%$ need help in English then 63 plus 34 or $97 \%$ need help in one or the other gives a number that is too large. The percentage that need help in both subjects must be subtracted off, else the people needing help in both are counted twice, once for needing help in mathematics and once again for needing help in English. The simple sum of the probabilities would work if the events in question were mutually exclusive, for then $P(A \cap B)$ is zero, and makes no difference.

EXAMPLE 19

Volunteers for a disaster relief effort were classified according to both specialty ( $C:$ construction, $E$ : education, $M$ : medicine) and language ability ( $S$ : speaks a single language fluently, $T$ : speaks two or more languages fluently). The results are shown in the following two-way classification table:

The first row of numbers means that 12 volunteers whose specialty is construction speak a single language fluently, and 1 volunteer whose specialty is construction speaks at least two languages fluently. Similarly for the other two rows.

A volunteer is selected at random, meaning that each one has an equal chance of being chosen. Find the probability that:

Solution:

When information is presented in a two-way classification table it is typically convenient to adjoin to the table the row and column totals, to produce a new table like this:

a. The probability sought is $P(M \cap T)$. The table shows that there are 2 such people, out of 28 in all, hence $P(M \cap T)=2 / 28 \approx 0.07$ or about a 7% chance.

b. The probability sought is $P(M \cup T)$. The third row total and the grand total in the sample give $P(M)=8 / 28$. The second column total and the grand total give $P(T)=6 / 28$. Thus using the result from part (a),

$\begin{align*} P(M \cup T)=P(M)+P(T)-P(M \cap T)=\frac{8}{28}+\frac{6}{28}-\frac{2}{28}=\frac{12}{28} \approx 0.43 \end{align*}$

or about a 43% chance.

c. This probability can be computed in two ways. Since the event of interest can be viewed as the event $C \cup E$ and the events $C$ and $E$ are mutually exclusive, the answer is, using the first two row totals,

$\begin{align*} P(C \cup E)=P(C)+P(E)-P(C \cap E)=\frac{13}{28}+\frac{7}{28}-\frac{0}{28}=\frac{20}{28} \approx 0.71 \end{align*}$

On the other hand, the event of interest can be thought of as the complement $M^{c}$ of $M$, hence using the value of $P(M)$ computed in part (b),

$\begin{align*} P\left(M^{c}\right)=1-P(M)=1-\frac{8}{28}=\frac{20}{28} \approx 0.71 \end{align*}$

as before.