Some events can be naturally expressed in terms of other, sometimes simpler, events.
Complements
Definition
The complement of an eventThe event does not occur. A in a sample space S, denoted A^{c}, is the collection of all outcomes in S that are not elements of the set A. It corresponds to negating any description in words of the event A.
Example 10
Two events connected with the experiment of rolling a single die are E: “the number rolled is even” and T: “the number rolled is greater than two.” Find the complement of each.
Solution:
In the sample space $S=\left\{\mathrm{1,2,3,4,5,6}\right\}$ the corresponding sets of outcomes are $E=\left\{\mathrm{2,4,6}\right\}$ and $T=\left\{\mathrm{3,4,5,6}\right\}.$ The complements are ${E}^{c}=\left\{\mathrm{1,3,5}\right\}$ and ${T}^{c}=\left\{\mathrm{1,2}\right\}.$
In words the complements are described by “the number rolled is not even” and “the number rolled is not greater than two.” Of course easier descriptions would be “the number rolled is odd” and “the number rolled is less than three.”
If there is a 60% chance of rain tomorrow, what is the probability of fair weather? The obvious answer, 40%, is an instance of the following general rule.
Probability Rule for Complements
$$P\left({A}^{c}\right)=1P\left(A\right)$$
This formula is particularly useful when finding the probability of an event directly is difficult.
Example 11
Find the probability that at least one heads will appear in five tosses of a fair coin.
Solution:
Identify outcomes by lists of five hs and ts, such as $tthtt$ and $hhttt.$ Although it is tedious to list them all, it is not difficult to count them. Think of using a tree diagram to do so. There are two choices for the first toss. For each of these there are two choices for the second toss, hence $2\times 2=4$ outcomes for two tosses. For each of these four outcomes, there are two possibilities for the third toss, hence $4\times 2=8$ outcomes for three tosses. Similarly, there are $8\times 2=16$ outcomes for four tosses and finally $16\times 2=32$ outcomes for five tosses.
Let O denote the event “at least one heads.” There are many ways to obtain at least one heads, but only one way to fail to do so: all tails. Thus although it is difficult to list all the outcomes that form O, it is easy to write ${O}^{c}=\left\{ttttt\right\}.$ Since there are 32 equally likely outcomes, each has probability 1/32, so $P\left({O}^{c}\right)=1\u221532$, hence $P\left(O\right)=11\u221532\approx 0.97$ or about a 97% chance.
Intersection of Events
Definition
The intersection of eventsBoth events occur. 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 lensshaped 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 ∩ T of the events E: “the number rolled is even” and T: “the number rolled is greater than two.”
Solution:
The sample space is $S=\left\{\mathrm{1,2,3,4,5,6}\right\}.$ Since the outcomes that are common to $E=\left\{\mathrm{2,4,6}\right\}$ and $T=\left\{\mathrm{3,4,5,6}\right\}$ are 4 and 6, $E\cap T=\left\{\mathrm{4,6}\right\}.$
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 ∩ T given above.
Example 13
A single die is rolled.
 Suppose the die is fair. Find the probability that the number rolled is both even and greater than two.
 Suppose the die has been “loaded” so that $P\left(1\right)=1\u221512$, $P\left(6\right)=3\u221512$, 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 $S=\left\{\mathrm{1,2,3,4,5,6}\right\}$ and the event in question is the intersection $E\cap T=\left\{\mathrm{4,6}\right\}$ of the previous example.
 Since the die is fair, all outcomes are equally likely, so by counting we have $P\left(E\cap T\right)=2\u22156.$

The information on the probabilities of the six outcomes that we have so far is
$$\begin{array}{ccccccc}\text{Outcome}\hfill & \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill & \hfill 5\hfill & \hfill 6\hfill \\ \text{Probablity}\hfill & \hfill \frac{1}{12}\hfill & \hfill p\hfill & \hfill p\hfill & \hfill p\hfill & \hfill p\hfill & \hfill \frac{3}{12}\hfill \end{array}$$
Since $P\left(1\right)+P\left(6\right)=4\u221512=1\u22153$ and the probabilities of all six outcomes add up to 1,
$$P\left(2\right)+P\left(3\right)+P\left(4\right)+P\left(5\right)=1\frac{1}{3}=\frac{2}{3}$$
Thus $4p=2\u22153$, so $p=1\u22156.$ In particular $P\left(4\right)=1\u22156.$ Therefore
$$P\left(E\cap T\right)=P\left(4\right)+P\left(6\right)=\frac{1}{6}+\frac{3}{12}=\frac{5}{12}$$
Definition
Events A and B are mutually exclusiveEvents that cannot both occur at once. 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
$$P\left(A\cap B\right)=0$$
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 $E=\left\{\mathrm{2,4,6}\right\}$ 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^{c}, the odds), {1,3}, and {5}.
Union of Events
Definition
The union of eventsOne or the other event occurs. A and B, denoted A ∪ 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 ∪ 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 $E=\left\{\mathrm{2,4,6}\right\}$ or $T=\left\{\mathrm{3,4,5,6}\right\}$ (or both) are 2, 3, 4, 5, and 6, $E\cup T=\left\{\mathrm{2,3,4,5,6}\right\}.$ 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 ∪ T given above.
Example 16
A twochild 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 ∪ D and $B\cup M.$
Solution:
A sample space for this experiment is $S=\left\{bb,bg,gb,gg\right\}$, 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
$$B=\left\{bb,bg,gb\right\}\text{\hspace{1em}}D=\left\{bg,gb\right\}\text{\hspace{1em}}M=\left\{bb,gg\right\}$$
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: $B\cup D=\left\{bb,bg,gb\right\}=B.$
Every outcome in the whole sample space S is in at least one or the other of the sets B and M, so $B\cup M=\left\{bb,bg,gb,gg\right\}=S.$
The following Additive Rule of Probability is a useful formula for calculating the probability of $A\cup B.$
Additive Rule of Probability
$$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)P\left(A\cap B\right)$$
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:
 both dice show a four
 at least one die shows a four
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.
$$\begin{array}{cccccc}11& 12& 13& 14& 15& 16\\ 21& 22& 23& 24& 25& 26\\ 31& 32& 33& 34& 35& 36\\ 41& 42& 43& 44& 45& 46\\ 51& 52& 53& 54& 55& 56\\ 61& 62& 63& 64& 65& 66\end{array}$$
 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.

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 $P\left({A}_{G}\right)=6\u221536$ and $P\left({A}_{R}\right)=6\u221536.$ Since ${A}_{G}\cap {A}_{R}=\left\{44\right\}$, $P\left({A}_{G}\cap {A}_{R}\right)=1\u221536$; this is the computation in part (a), of course. Thus by the Additive Rule of Probability,
$$P({A}_{G}\cup {A}_{R})=P({A}_{G})+P({A}_{R})P({A}_{G}{A}_{R})=\frac{6}{36}+\frac{6}{36}\frac{1}{36}=\frac{11}{36}$$
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 $P\left(M\right)=0.63$, $P\left(E\right)=0.34$, and $P\left(M\cap E\right)=0.27.$ The Additive Rule of Probability gives
$$P\left(M\cup E\right)=P\left(M\right)+P\left(E\right)P\left(M\cap E\right)=0.63+0.340.27=0.70$$
Note how the naïve 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 $P\left(A\cap B\right)$ 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 twoway classification table:
Specialty  Language Ability  

S  T  
C  12  1 
E  4  3 
M  6  2 
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:
 his specialty is medicine and he speaks two or more languages;
 either his specialty is medicine or he speaks two or more languages;
 his specialty is something other than medicine.
Solution:
When information is presented in a twoway classification table it is typically convenient to adjoin to the table the row and column totals, to produce a new table like this:
Specialty  Language Ability  Total  

S  T  
C  12  1  13 
E  4  3  7 
M  6  2  8 
Total  22  6  28 
 The probability sought is $P\left(M\cap T\right).$ The table shows that there are 2 such people, out of 28 in all, hence $P\left(M\cap T\right)=2\u221528\approx 0.07$ or about a 7% chance.

The probability sought is $P\left(M\cup T\right).$ The third row total and the grand total in the sample give $P\left(M\right)=8\u221528.$ The second column total and the grand total give $P\left(T\right)=6\u221528.$ Thus using the result from part (a),
$$P\left(M\cup T\right)=P\left(M\right)+P\left(T\right)P\left(M\cap T\right)=\frac{8}{28}+\frac{6}{28}\frac{2}{28}=\frac{12}{28}\approx 0.43$$or about a 43% chance.

This probability can be computed in two ways. Since the event of interest can be viewed as the event C ∪ E and the events C and E are mutually exclusive, the answer is, using the first two row totals,
$$P\left(C\cup E\right)=P\left(C\right)+P\left(E\right)P\left(C\cap E\right)=\frac{13}{28}+\frac{7}{28}\frac{0}{28}=\frac{20}{28}\approx 0.71$$On the other hand, the event of interest can be thought of as the complement M^{c} of M, hence using the value of $P\left(M\right)$ computed in part (b),
$$P\left({M}^{c}\right)=1P\left(M\right)=1\frac{8}{28}=\frac{20}{28}\approx 0.71$$as before.
Key Takeaway
 The probability of an event that is a complement or union of events of known probability can be computed using formulas.
Exercises

For the sample space $S=\left\{a,b,c,d,e\right\}$ identify the complement of each event given.
 $A=\left\{a,d,e\right\}$
 $B=\left\{b,c,d,e\right\}$
 S

For the sample space $S=\left\{r,s,t,u,v\right\}$ identify the complement of each event given.
 $R=\left\{t,u\right\}$
 $T=\left\{r\right\}$
 ∅ (the “empty” set that has no elements)

The sample space for three tosses of a coin is
$$S=\left\{hhh,hht,hth,htt,thh,tht,tth,ttt\right\}$$
Define events
$$\begin{array}{c}H:\mathrm{\text{at\hspace{0.17em}least\hspace{0.17em}one\hspace{0.17em}head\hspace{0.17em}is\hspace{0.17em}observed}}\hfill \\ M:\mathrm{\text{more\hspace{0.17em}heads\hspace{0.17em}than\hspace{0.17em}tails\hspace{0.17em}are\hspace{0.17em}observed}}\hfill \end{array}$$
 List the outcomes that comprise H and M.
 List the outcomes that comprise H ∩ M, H ∪ M, and H^{c}.
 Assuming all outcomes are equally likely, find $P\left(H\cap M\right)$, $P\left(H\cup M\right)$, and $P\left({H}^{c}\right).$
 Determine whether or not H^{c} and M are mutually exclusive. Explain why or why not.

For the experiment of rolling a single sixsided die once, define events
$$\begin{array}{c}T:\mathrm{\text{the\hspace{0.17em}number\hspace{0.17em}rolled\hspace{0.17em}is\hspace{0.17em}three}}\hfill \\ G:\mathrm{\text{the\hspace{0.17em}number\hspace{0.17em}rolled\hspace{0.17em}is\hspace{0.17em}four\hspace{0.17em}or\hspace{0.17em}greater}}\hfill \end{array}$$
 List the outcomes that comprise T and G.
 List the outcomes that comprise T ∩ G, T ∪ G, T^{c}, and ${\left(T\cup G\right)}^{c}.$
 Assuming all outcomes are equally likely, find $P\left(T\cap G\right)$, $P\left(T\cup G\right)$, and $P\left({T}^{c}\right).$
 Determine whether or not T and G are mutually exclusive. Explain why or why not.

A special deck of 16 cards has 4 that are blue, 4 yellow, 4 green, and 4 red. The four cards of each color are numbered from one to four. A single card is drawn at random. Define events
$$\begin{array}{c}B:\mathrm{\text{the\hspace{0.17em}card\hspace{0.17em}is\hspace{0.17em}blue}}\hfill \\ R:\mathrm{\text{the\hspace{0.17em}card\hspace{0.17em}is\hspace{0.17em}red}}\hfill \\ N:\mathrm{\text{the\hspace{0.17em}number\hspace{0.17em}on\hspace{0.17em}the\hspace{0.17em}card\hspace{0.17em}is\hspace{0.17em}at\hspace{0.17em}most\hspace{0.17em}two}}\hfill \end{array}$$
(Video) Intersection of Sets, Union of Sets and Venn Diagrams List the outcomes that comprise B, R, and N.
 List the outcomes that comprise B ∩ R, B ∪ R, B ∩ N, R ∪ N, B^{c}, and ${\left(B\cup R\right)}^{c}.$
 Assuming all outcomes are equally likely, find the probabilities of the events in the previous part.
 Determine whether or not B and N are mutually exclusive. Explain why or why not.

In the context of the previous problem, define events
$$\begin{array}{c}Y:\mathrm{\text{the\hspace{0.17em}card\hspace{0.17em}is\hspace{0.17em}yellow}}\hfill \\ I:\mathrm{\text{the\hspace{0.17em}number\hspace{0.17em}on\hspace{0.17em}the\hspace{0.17em}card\hspace{0.17em}is\hspace{0.17em}not\hspace{0.17em}a\hspace{0.17em}one}}\hfill \\ J:\mathrm{\text{the\hspace{0.17em}number\hspace{0.17em}on\hspace{0.17em}the\hspace{0.17em}card\hspace{0.17em}is\hspace{0.17em}a\hspace{0.17em}two\hspace{0.17em}or\hspace{0.17em}a\hspace{0.17em}four}}\hfill \end{array}$$
 List the outcomes that comprise Y, I, and J.
 List the outcomes that comprise Y ∩ I, Y ∪ J, I ∩ J, I^{c}, and ${\left(Y\cup J\right)}^{c}.$
 Assuming all outcomes are equally likely, find the probabilities of the events in the previous part.
 Determine whether or not I^{c} and J are mutually exclusive. Explain why or why not.

The Venn diagram provided shows a sample space and two events A and B. Suppose $P\left(a\right)=0.13$, $P\left(b\right)=0.09$, $P\left(c\right)=0.27$, $P\left(d\right)=0.20$, and $P\left(e\right)=0.31.$ Confirm that the probabilities of the outcomes add up to 1, then compute the following probabilities.
 $P\left(A\right).$
 $P\left(B\right).$
 $P\left({A}^{c}\right)$ two ways: (i) by finding the outcomes in A^{c} and adding their probabilities, and (ii) using the Probability Rule for Complements.
 $P\left(A\cap B\right).$
 $P\left(A\cup B\right)$ two ways: (i) by finding the outcomes in A ∪ B and adding their probabilities, and (ii) using the Additive Rule of Probability.

The Venn diagram provided shows a sample space and two events A and B. Suppose $P\left(a\right)=0.32$, $P\left(b\right)=0.17$, $P\left(c\right)=0.28$, and $P\left(d\right)=0.23.$ Confirm that the probabilities of the outcomes add up to 1, then compute the following probabilities.
 $P\left(A\right).$
 $P\left(B\right).$
 $P\left({A}^{c}\right)$ two ways: (i) by finding the outcomes in A^{c} and adding their probabilities, and (ii) using the Probability Rule for Complements.
 $P\left(A\cap B\right).$
 $P\left(A\cup B\right)$ two ways: (i) by finding the outcomes in A ∪ B and adding their probabilities, and (ii) using the Additive Rule of Probability.

Confirm that the probabilities in the twoway contingency table add up to 1, then use it to find the probabilities of the events indicated.
U V W A 0.15 0.00 0.23 B 0.22 0.30 0.10  $P\left(A\right)$, $P\left(B\right)$, $P\left(A\cap B\right).$
 $P\left(U\right)$, $P\left(W\right)$, $P\left(U\cap W\right).$
 $P\left(U\cup W\right).$
 $P\left({V}^{c}\right).$
 Determine whether or not the events A and U are mutually exclusive; the events A and V.

Confirm that the probabilities in the twoway contingency table add up to 1, then use it to find the probabilities of the events indicated.
R S T M 0.09 0.25 0.19 N 0.31 0.16 0.00  $P\left(R\right)$, $P\left(S\right)$, $P\left(R\cap S\right).$
 $P\left(M\right)$, $P\left(N\right)$, $P\left(M\cap N\right).$
 $P\left(R\cup S\right).$
 $P\left({R}^{c}\right).$
 Determine whether or not the events N and S are mutually exclusive; the events N and T.
Basic

Make a statement in ordinary English that describes the complement of each event (do not simply insert the word “not”).
 In the roll of a die: “five or more.”
 In a roll of a die: “an even number.”
 In two tosses of a coin: “at least one heads.”
 In the random selection of a college student: “Not a freshman.”

Make a statement in ordinary English that describes the complement of each event (do not simply insert the word “not”).
 In the roll of a die: “two or less.”
 In the roll of a die: “one, three, or four.”
 In two tosses of a coin: “at most one heads.”
 In the random selection of a college student: “Neither a freshman nor a senior.”

The sample space that describes all threechild families according to the genders of the children with respect to birth order is
$$S=\left\{bbb,bbg,bgb,bgg,gbb,gbg,ggb,ggg\right\}.$$
For each of the following events in the experiment of selecting a threechild family at random, state the complement of the event in the simplest possible terms, then find the outcomes that comprise the event and its complement.
 At least one child is a girl.
 At most one child is a girl.
 All of the children are girls.
 Exactly two of the children are girls.
 The first born is a girl.

The sample space that describes the twoway classification of citizens according to gender and opinion on a political issue is
$$S=\left\{mf,ma,mn,ff,fa,fn\right\},$$
where the first letter denotes gender (m: male, f: female) and the second opinion (f: for, a: against, n: neutral). For each of the following events in the experiment of selecting a citizen at random, state the complement of the event in the simplest possible terms, then find the outcomes that comprise the event and its complement.
 The person is male.
 The person is not in favor.
 The person is either male or in favor.
 The person is female and neutral.

A tourist who speaks English and German but no other language visits a region of Slovenia. If 35% of the residents speak English, 15% speak German, and 3% speak both English and German, what is the probability that the tourist will be able to talk with a randomly encountered resident of the region?

In a certain country 43% of all automobiles have airbags, 27% have antilock brakes, and 13% have both. What is the probability that a randomly selected vehicle will have both airbags and antilock brakes?

A manufacturer examines its records over the last year on a component part received from outside suppliers. The breakdown on source (supplier A, supplier B) and quality (H: high, U: usable, D: defective) is shown in the twoway contingency table.
H U D A 0.6937 0.0049 0.0014 B 0.2982 0.0009 0.0009 The record of a part is selected at random. Find the probability of each of the following events.
 The part was defective.
 The part was either of high quality or was at least usable, in two ways: (i) by adding numbers in the table, and (ii) using the answer to (a) and the Probability Rule for Complements.
 The part was defective and came from supplier B.
 The part was defective or came from supplier B, in two ways: by finding the cells in the table that correspond to this event and adding their probabilities, and (ii) using the Additive Rule of Probability.

Individuals with a particular medical condition were classified according to the presence (T) or absence (N) of a potential toxin in their blood and the onset of the condition (E: early, M: midrange, L: late). The breakdown according to this classification is shown in the twoway contingency table.
E M L T 0.012 0.124 0.013 N 0.170 0.638 0.043 One of these individuals is selected at random. Find the probability of each of the following events.
 The person experienced early onset of the condition.
 The onset of the condition was either midrange or late, in two ways: (i) by adding numbers in the table, and (ii) using the answer to (a) and the Probability Rule for Complements.
 The toxin is present in the person’s blood.
 The person experienced early onset of the condition and the toxin is present in the person’s blood.
 The person experienced early onset of the condition or the toxin is present in the person’s blood, in two ways: (i) by finding the cells in the table that correspond to this event and adding their probabilities, and (ii) using the Additive Rule of Probability.

The breakdown of the students enrolled in a university course by class (F: freshman, $So$: sophomore, J: junior, $Se$: senior) and academic major (S: science, mathematics, or engineering, L: liberal arts, O: other) is shown in the twoway classification table.
Major Class F So J Se S 92 42 20 13 L 368 167 80 53 O 460 209 100 67 A student enrolled in the course is selected at random. Adjoin the row and column totals to the table and use the expanded table to find the probability of each of the following events.
 The student is a freshman.
 The student is a liberal arts major.
 The student is a freshman liberal arts major.
 The student is either a freshman or a liberal arts major.
 The student is not a liberal arts major.
(Video) Union, Intersection, and Complement: Everything you need to know! 
The table relates the response to a fundraising appeal by a college to its alumni to the number of years since graduation.
Response Years Since Graduation 0–5 6–20 21–35 Over 35 Positive 120 440 210 90 None 1380 3560 3290 910 An alumnus is selected at random. Adjoin the row and column totals to the table and use the expanded table to find the probability of each of the following events.
 The alumnus responded.
 The alumnus did not respond.
 The alumnus graduated at least 21 years ago.
 The alumnus graduated at least 21 years ago and responded.
Applications

The sample space for tossing three coins is
$$S=\left\{hhh,hht,hth,htt,thh,tht,tth,ttt\right\}$$
 List the outcomes that correspond to the statement “All the coins are heads.”
 List the outcomes that correspond to the statement “Not all the coins are heads.”
 List the outcomes that correspond to the statement “All the coins are not heads.”
Additional Exercises
Answers

 $\left\{b,c\right\}$
 $\left\{a\right\}$
 ∅

 $H=\left\{hhh,hht,hth,htt,thh,tht,tth\right\}$, $M=\left\{hhh,hht,hth,thh\right\}$
 $H\cap M=\left\{hhh,hht,hth,thh\right\}$, $H\cup M=H$, ${H}^{c}=\left\{ttt\right\}$
 $P\left(H\cap M\right)=4\u22158$, $P\left(H\cup M\right)=7\u22158$, $P\left({H}^{c}\right)=1\u22158$
 Mutually exclusive because they have no elements in common.

 $B=\left\{b1,b2,b3,b4\right\}$, $R=\left\{r1,r2,r3,r4\right\}$, $N=\left\{b1,b2,y1,y2,g1,g2,r1,r2\right\}$
 $B\cap R=\varnothing $, $B\cup R=\left\{b1,b2,b3,b4,r1,r2,r3,r4\right\},$ $B\cap N=\left\{b1,b2\right\}$, $R\cup N=\left\{b1,b2,y1,y2,g1,g2,r1,r2,r3,r4\right\},$ ${B}^{c}=\left\{y1,y2,y3,y4,g1,g2,g3,g4,r1,r2,r3,r4\right\},$ ${\left(B\cup R\right)}^{c}=\left\{y1,y2,y3,y4,g1,g2,g3,g4\right\}$
 $P\left(B\cap R\right)=0$, $P\left(B\cup R\right)=8\u221516$, $P\left(B\cap N\right)=2\u221516$, $P\left(R\cup N\right)=10\u221516$, $P\left({B}^{c}\right)=12\u221516$, $P\left({\left(B\cup R\right)}^{c}\right)=8\u221516$
 Not mutually exclusive because they have an element in common.

 0.36
 0.78
 0.64
 0.27
 0.87

 $P\left(A\right)=0.38$, $P\left(B\right)=0.62$, $P\left(A\cap B\right)=0$
 $P\left(U\right)=0.37$, $P\left(W\right)=0.33$, $P\left(U\cap W\right)=0$
 0.7
 0.7
 A and U are not mutually exclusive because $P\left(A\cap U\right)$ is the nonzero number 0.15. A and V are mutually exclusive because $P\left(A\cap V\right)=0.$

 “four or less”
 “an odd number”
 “no heads” or “all tails”
 “a freshman”


“All the children are boys.”
Event: $\left\{bbg,bgb,bgg,gbb,gbg,ggb,ggg\right\}$,
Complement: $\left\{bbb\right\}$

“At least two of the children are girls” or “There are two or three girls.”
Event: $\left\{bbb,bbg,bgb,gbb\right\}$,
Complement: $\left\{bgg,gbg,ggb,ggg\right\}$

“At least one child is a boy.”
Event: $\left\{ggg\right\}$,
Complement: $\left\{bbb,bbg,bgb,bgg,gbb,gbg,ggb\right\}$

“There are either no girls, exactly one girl, or three girls.”
Event: $\left\{bgg,gbg,ggb\right\}$,
Complement: $\left\{bbb,bbg,bgb,gbb,ggg\right\}$

“The first born is a boy.”
Event: $\left\{gbb,gbg,ggb,ggg\right\}$,
Complement: $\left\{bbb,bbg,bgb,bgg\right\}$


0.47

 0.0023
 0.9977
 0.0009
 0.3014

 920/1671
 668/1671
 368/1671
 1220/1671
 1003/1671

 $\left\{hhh\right\}$
 $\left\{hht,hth,htt,thh,tht,tth,ttt\right\}$
 $\left\{ttt\right\}$
FAQs
What are complements of unions and intersections? ›
The union is notated A ⋃ B. The intersection of two sets contains only the elements that are in both sets. The intersection is notated A ⋂ B. The complement of a set A contains everything that is not in the set A.
What is the answer of a union a complement? ›A complement union B complement can be understood as the union of the complements of each of the two sets. The union of the complement of set A and set B is equal to the difference of the universal set(μ) and the intersection of the two sets (A n B).
What is the answer of A∩ B? ›Case II: (A ∩ B)' = A' U B'
Answer: In mathematics, the intersection of two given sets is the largest set that contains all the elements that are common to both the sets. In addition, the symbol for denoting intersection of sets is ∩, which is a common representation of sets.
A union B complement is a formula in set theory that is equal to the intersection of the complements of the sets A and B. Mathematically, the formula for A union B Complement is given by, (A U B)' = A' ∩ B' or (A U B)^{c} = A^{c} ∩ B^{c}, where ' or ^{c} denote the complement of a set.
What is an example of intersection complement? ›A Intersection B Complement Examples
Example 2: Consider U = {1, 2, 3, 4, 5, 6, 7, 8, 9} and A' = {2, 3, 5} and B' = {1, 2, 3, 4, 5}. Find the elements in A Intersection B Complement. Answer: The elements in A Intersection B Complement are {1, 2, 3, 4, 5}.
The complement of set A is defined as a set that contains the elements present in the universal set but not in set A. For example, Set U = {2, 4, 6, 8, 10, 12} and set A = {4, 6, 8}, then the complement of set A, A′ = {2, 10, 12}.
What does ∩ and ∪ mean in math? ›∪: Union of two sets. A complete Venn diagram represents the union of two sets. ∩: Intersection of two sets. The intersection shows what items are shared between categories.
What is the complement of A ∪ B? ›De Morgan's Laws:
The complement of the union of two sets is the intersection of the complements of the individual sets: ( A ∪ B ) ′ = A ′ ∩ B ′ .
For a union and an intersection example, use the set B = {1, 2, 3, 4, 5, 6} and the set D = {3, 5, 7, 9, 10}. The union of sets B and D is the set {1, 2, 3, 4, 5, 6, 7, 9, 10}. The intersection of sets B and D is the set {3, 5}.
What does PA ∩ B ') mean? ›Joint probability: p(A ∩B). Joint probability is that of event A and event B occurring. It is the probability of the intersection of two or more events. The probability of the intersection of A and B may be written p(A ∩ B).
How do you calculate A∩ B? ›
FAQs on A∩B Formula
P(A∩B) is the probability of both independent events “A” and "B" happening together, P(A∩B) formula can be written as P(A∩B) = P(A) × P(B), where, P(A∩B) = Probability of both independent events “A” and "B" happening together.
The union of two sets A and B is a set that contains all the elements of A and B and is denoted by A U B (which can be read as "A or B" (or) "A union B").
How do you find the complement intersection of a B complement? ›What Is A Complement Intersection B Complement? The set A complement intersection B complement can be obtained by taking the common elements of the complement of set A, and the complement of set B. This set can also be obtained after removing the union of the two sets from the universal set. A' n B' = μ  (A U B).
How do you remember unions and intersections? ›intersection and union symbols
The intersection ∪ and union ∩ symbols look a little like letters in the alphabet. In fact, that's a trick for remembering them. The union symbol looks like a capital U, for union.
To find the intersection of two or more sets, you look for elements that are contained in all of the sets. To find the union of two or more sets, you combine all the elements from each set together, making sure to remove any duplicates.
What is the difference between union and intersect? ›The UNION operation combines the results of two subqueries into a single result that comprises the rows that are returned by both queries. The INTERSECT operation combines the results of two queries into a single result that comprises all the rows common to both queries.
What is a intersection B with example? ›For any two sets A and B, the intersection, A ∩ B (read as A intersection B) lists all the elements that are present in both sets (common elements of A and B). For example, if Set A = {1,2,3,4,5} and Set B = {3,4,6,8}, A ∩ B = {3,4}.
How do you find the complement in a sentence? ›A subject complement is found in the predicate of a sentence (the part of the sentence that contains the verb and makes a statement about the subject). The subject complement follows a linking verb (a verb that expresses a state of being).
How do you solve a union? ›Consider two sets, A and B, such that the number of elements in the union of A and B can be calculated as follows. n(A U B) = n(A) + n(B) – n(A ∩ B)
What is union of two sets? ›The union of two sets A and B is the set of elements which are in A, in B, or in both A and B. In setbuilder notation, . For example, if A = {1, 3, 5, 7} and B = {1, 2, 4, 6, 7} then A ∪ B = {1, 2, 3, 4, 5, 6, 7}.
What is a complement in math? ›
In general, the word "complement" refers to that subset of some set which excludes a given subset . Taking and its complement together then gives the whole of the original set. The notations and are commonly used to denote the complement of a set .
What is one example of a union? ›Nearly every profession, industry, and organization have its unions. Examples include nurses' unions, teachers' unions, drivers', labor unions, and lawyers' unions. They don't work under management, but unions have the power to accept or reject any terms and conditions on behalf of workers.
Is union multiplication or addition? ›Union and intersection
This second formula is the same addition rule calculation, but we use the ∪ and ∩ symbols instead of the words “and” and “or.” P ( A ∪ B ) P(A\cup B) P(A∪B) is called the union of A and B, and it means the probability of either A or B or both occurring.
The Union is any region including either A or B. The Intersection is any region including both A and B. The diagrams we have drawn are called the Venn diagrams.
What is the probability of AUB? ›P(A U B) is the probability of the sum of all sample points in A U B. Now P(A) + P(B) is the sum of probabilities of sample points in A and in B. Since we added up the sample points in (A ∩ B) twice, we need to subtract once to obtain the sum of probabilities in (A U B), which is P(A U B).
What does aub mean? ›Abnormal uterine bleeding (AUB) is bleeding from the uterus that is longer than usual or that occurs at an irregular time. Bleeding may be heavier or lighter than usual and occur often or randomly. AUB can occur: As spotting or bleeding between your periods. After sex.
What is the formula for A or B? ›The rule for finding the probability of either/or problems, we need to think about the possibility of one or more outcomes happening together. The formula for finding the either/or probability is P(A or B) = P(A) +P(B)  P (A and B).
How do you find the probability of A or B but not both? ›The Probability Definition of an Event
Probability of event 'A or B' = P(A ∪ B) = P(A) + P(B) – P(A ∩ B). The probability of event 'A but not B' = P(A ∩ B') = P(A) – P(A ∩ B)
P(A and B) = P(A)P(B). The chance of all of two or more events occurring is called the intersection of events. For independent events, the probability of the intersection of two or more events is the product of the probabilities.
How do you find the intersection of three events? ›To calculate the probability of the intersection of more than two events, the conditional probabilities of all of the preceding events must be considered. In the case of three events, A, B, and C, the probability of the intersection P(A and B and C) = P(A)P(BA)P(CA and B).
Is 0 an integer? ›
As a whole number that can be written without a remainder, 0 classifies as an integer.
What is a set Grade 2? ›Sets, in mathematics, are an organized collection of objects and can be represented in setbuilder form or roster form. Usually, sets are represented in curly braces {}, for example, A = {1,2,3,4} is a set.
What is the union formula for 3 sets? ›Union of three events (inclusion/exclusion formula): P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(A ∩ C) − P(B ∩ C) + P(A ∩ B ∩ C). Draw Venn diagrams: Venn diagrams help you picture what is going on and deriving the appropriate probabilities.
What is the formula to calculate a intersection B intersection c? ›Here we first find A n B and B n C, and finally, we get the answer for A n B n C. Hence we have the formula A n B n C = (A n B) n ( B n C). Let us learn more about A intersection B intersection C, and how to find A intersection B intersection C, with the help of examples, FAQs.
What is the U and upside down U in math? ›What is U and upside down U in math? In math, the symbol U represents the union of two sets, while upsidedown U represents the intersection of the sets.
What does this symbol mean ∩? ›∩ The symbol ∩ means intersection. Given two sets S and T, S ∩ T is used to denote the set {xx ∈ S and x ∈ T}. For example {1,2,3}∩{3,4,5} = {3}.
What does C mean in math? ›The Latin small letter c is used in math to represent a variable or coefficient.
What is the complement of intersection probability? ›This means that the complement of their intersection is the entire sample space, which has a probability of one. Or more formally, we can say that the probability of 𝐴 intersect 𝐵 complement is one minus the probability of 𝐴 intersect 𝐵. That's one minus zero, which is one.
What is the relationship between union and intersection? ›The union function of two sets has all the elements or objects present in two sets or either of the two sets. It is represented by ⋃. The intersection function of two sets is when all the elements present in the both sets are present. It is represented as ⋂.
Is the complement of the union of two sets the same as the intersection of the complements of each set? ›(a) The Complement of the union of two sets is the intersection of their complements and the complement of the intersection of two sets is the union of their complements. These are called De Morgan's Law.
What is the complement of the universal set? ›
Complement of Universal Set
The empty set is defined as the complement of the universal set. That means where Universal set consists of a set of all elements, the empty set contains no elements of the subsets. The empty set is also called a Null set and is denoted by '{}'.
When one of two disjoint events must occur, the two events are said to be complementary. Since one or the other event must occur, the sum of the probabilities of the two complementary events adds up to 1, or 100 percent of the outcomes of the events.
How do you find the complement of a probability? ›The complement, AC , of an event A consists of all of the outcomes in the sample space that are NOT in event A . The probability of the complement can be found from the original event using the formula: P(AC)=1−P(A) P ( A C ) = 1 − P ( A ) .
What is the probability formula for union and intersection? ›Intersection and unions are useful to assess the probability of two events occurring together and the probability of at least one of the two events. P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
How do you calculate union and intersection in probability? ›If two events A and B are not disjoint, then the probability of their union (the event that A or B occurs) is equal to the sum of their probabilities minus the sum of their intersection. P(A) + P(B)  P(A and B) = 3/5 + 2/5  6/25 = 1  6/25 = 19/25 = 0.76.
What is the difference between union of two sets and intersection of two sets? ›What is the difference between union and intersection? A union of sets produces a new set containing each element present in the original sets. An intersection of sets produces a new set that contains only the elements that the original sets have in common.
What is the formula for union and intersection of two sets? ›= n(A) + n(B) – n(A ∩ B) Simply, the number of elements in the union of set A and B is equal to the sum of cardinal numbers of the sets A and B, minus that of their intersection.
Is the complement of two sets the difference of two sets? ›Complement and Difference of Sets
The complement of a set A is denoted by A' or A^{c} and it is the difference of the sets U and A, where U is the universal set. i.e., A' (or) A^{c} = U  A. This refers to the set of all elements that are in the universal set that are not elements of set A.
Answer and Explanation: The complement of the universal set is the null set. While the universal set contains all the possible elements, the null set contains no elements at all.
What is the difference between universal set and complementary set? ›A universal set is a set that contains all the elements we are interested in. This would have to be defined by the context. A complement is relative to the universal set, so A^{c} contains all the elements in the universal set that are not in A.
Is the complement of a set unique? ›
So complements are unique.