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Unit 2: Probability, Random Variables, and Probability Distributions

Mutually Exclusive Events

Find the probability that two events both happen, read it from a table or a description, and use it to justify whether two events are mutually exclusive.

Topic 2.5 · 8 min read

Roll one six-sided die. "The roll is even" and "the roll is a 5" can't both happen on that roll, because 5 is odd. "The roll is even" and "the roll is greater than 3" can both happen: a 4 or a 6 does it. Whether two events can happen together is a probability question, and the probability that answers it has a name.

Joint probability: both events at once

For two events AA and BB in the same random process, the event "AA and BB" is the set of outcomes that belong to both. It is called the intersection of AA and BB and written A∩BA \cap B. Its probability, P(A∩B)P(A \cap B), is the joint probability of AA and BB: the chance that one run of the process produces an outcome in both events.

With the die, let AA = even ={2,4,6}= \{2, 4, 6\} and CC = greater than 3 ={4,5,6}= \{4, 5, 6\}. The outcomes in both are {4,6}\{4, 6\}, and the six outcomes are equally likely (topic 2.4), so

P(A∩C)=26=13.P(A \cap C) = \frac{2}{6} = \frac{1}{3}.

The joint probability is never larger than either event's own probability, because A∩CA \cap C is part of AA and also part of CC. Here 13\frac{1}{3} is less than P(A)=12P(A) = \frac{1}{2} and less than P(C)=12P(C) = \frac{1}{2}.

Reading a joint probability from a two-way table

Most AP questions give the joint probability as a cell of a two-way table. A public library recorded the 500 items checked out on one Saturday by format and by the borrower's age group:

ChildAdultTotal
Print book140160300
Ebook30110140
Audiobook105060
Total180320500

Pick one of these 500 checkouts at random. The event "ebook and child" is the one cell where the Ebook row meets the Child column, so

P(ebook∩child)=30500=0.06.P(\text{ebook} \cap \text{child}) = \frac{30}{500} = 0.06.

The denominator is the grand total, 500, because the random selection is from all 500 checkouts. Two nearby numbers are different questions: 140500=0.28\frac{140}{500} = 0.28 is P(ebook)P(\text{ebook}), which ignores age, and 30180\frac{30}{180} is the proportion of children's checkouts that were ebooks, a conditional probability (topic 2.6).

Mutually exclusive events

Two events are mutually exclusive, also called disjoint, when they cannot both occur on the same run of the process. They share no outcomes, so their intersection is empty and

P(A∩B)=0.P(A \cap B) = 0.

The reverse also holds: if the joint probability is 0, the events never happen together, so they are mutually exclusive. That makes the joint probability the tool for justifying either answer.

In the library table, every checkout has exactly one format and exactly one age group. So "print book" and "ebook" are mutually exclusive, and so are "child" and "adult": the categories of one variable never overlap when each individual gets one value. Events built from two different variables are mutually exclusive only if their cell is 0. "Ebook" and "child" are not, because P(ebook∩child)=0.06>0P(\text{ebook} \cap \text{child}) = 0.06 > 0.

What a zero cell does and doesn't show

A zero in a table of data shows that among these individuals no one fell in both categories, so for a random pick from them the joint probability is exactly 0. For a larger population, it means you saw no overlap in this sample, which is evidence but not proof. When a question asks about the individuals in the table, the zero cell settles it. When the events can't overlap by their definitions (one die roll can't be both 2 and 5), they are mutually exclusive for every run, with or without data.

A quick test with two probabilities

If AA and BB are mutually exclusive, their outcomes sit in separate parts of the sample space, so together they use up P(A)+P(B)P(A) + P(B) of the total probability, which is 1. That gives a fast check: if P(A)+P(B)>1P(A) + P(B) > 1, the events cannot be mutually exclusive. With P(A)=0.6P(A) = 0.6 and P(B)=0.5P(B) = 0.5, the two events would need 1.1 of the probability if they didn't overlap, and only 1 exists, so at least 0.10.1 of the probability must be in both.

The test only works in one direction. P(A)+P(B)≤1P(A) + P(B) \le 1 does not show the events are mutually exclusive; it only fails to rule it out. The die events "even" and "greater than 3" have 12+12=1\frac{1}{2} + \frac{1}{2} = 1 and still overlap.

Mutually exclusive is a different idea from independent. If AA and BB are disjoint and you learn AA happened, you know for certain that BB did not, which is the opposite of the two events having nothing to do with each other. Topic 2.7 defines independence and shows why disjoint events with positive probabilities are never independent.

Worked examples

Worked exampleExample 1: a raffle drawing

Tickets numbered 1 through 40 go in a drum, and one is drawn at random. Let FF = the number is a multiple of 5, LL = the number is less than 5, and EE = the number is even. For each pair (FF and LL, FF and EE, LL and EE), find the joint probability and say whether the events are mutually exclusive.

Show solution

List each event first, because the overlaps are then visible. F={5,10,15,20,25,30,35,40}F = \{5, 10, 15, 20, 25, 30, 35, 40\} (8 numbers), L={1,2,3,4}L = \{1, 2, 3, 4\}, and EE is the 20 even numbers. All 40 tickets are equally likely.

FF and LL: the smallest multiple of 5 is 5, which is not less than 5, so no ticket is in both. P(F∩L)=0P(F \cap L) = 0, and FF and LL are mutually exclusive.

FF and EE: the even multiples of 5 are 10, 20, 30, 40. P(F∩E)=440=0.1P(F \cap E) = \frac{4}{40} = 0.1. Since this is greater than 0, FF and EE are not mutually exclusive.

LL and EE: tickets 2 and 4 are in both. P(L∩E)=240=0.05>0P(L \cap E) = \frac{2}{40} = 0.05 > 0, so they are not mutually exclusive.

Worked exampleExample 2: a veterinary clinic's records

A veterinary clinic classified its 320 visits in March by species and type of visit.

WellnessIllnessSurgeryTotal
Dog906030180
Cat504020110
Rabbit1614030
Total15611450320

One of these visits is selected at random. (a) Find P(cat∩surgery)P(\text{cat} \cap \text{surgery}). (b) Are the events "the visit is for a rabbit" and "the visit is a surgery" mutually exclusive? Justify. (c) Are "the visit is for a dog" and "the visit is for an illness" mutually exclusive? Justify.

Show solution

(a) The Cat row meets the Surgery column at 20, and the selection is from all 320 visits: P(cat∩surgery)=20320=0.0625P(\text{cat} \cap \text{surgery}) = \frac{20}{320} = 0.0625.

(b) Yes. None of the 320 visits was a rabbit surgery, so P(rabbit∩surgery)=0320=0P(\text{rabbit} \cap \text{surgery}) = \frac{0}{320} = 0. A joint probability of 0 means the two events cannot both happen for a visit chosen from these records, so they are mutually exclusive.

(c) No. P(dog∩illness)=60320=0.1875P(\text{dog} \cap \text{illness}) = \frac{60}{320} = 0.1875, which is greater than 0: 60 of the visits were dogs seen for an illness.

The answer to (b) is about these 320 visits. It would be wrong to conclude the clinic never operates on rabbits; it only did none in March.

Worked exampleExample 3: exam-style free response

A survey of the students at a large high school found that 58% have a part-time job and 47% play on a school sports team. Let JJ = a randomly selected student has a part-time job and SS = the student plays on a sports team.

(a) A classmate says JJ and SS might be mutually exclusive, because students with jobs have less time for sports. Using the given probabilities, explain why JJ and SS cannot be mutually exclusive.

(b) The survey also found that 14% of students both have a job and play on a team. Find the probability that a randomly selected student has a part-time job but does not play on a sports team.

Show solution

(a) If JJ and SS were mutually exclusive, no student would be in both, and the students with jobs and the students on teams would be separate groups making up 0.58+0.47=1.050.58 + 0.47 = 1.05 of the school. A probability can't exceed 1, so the groups must overlap: at least 1.05−1=0.051.05 - 1 = 0.05 of students have a job and play a sport. So P(J∩S)>0P(J \cap S) > 0, and JJ and SS are not mutually exclusive.

(b) The 58% with jobs split into those who also play a sport and those who don't. The first group is 14%, so

P(J∩SC)=P(J)−P(J∩S)=0.58−0.14=0.44.P(J \cap S^C) = P(J) - P(J \cap S) = 0.58 - 0.14 = 0.44.

The probability that a randomly selected student at this school has a part-time job but does not play on a sports team is 0.44.

Why these moves: (a) is a justification, so it names the joint probability and why it can't be 0, using the numbers given rather than the classmate's reasoning about time. In (b), the event "job and no sport" is one region of a Venn diagram, and subtracting the overlap from P(J)P(J) finds it without needing any other information.

Where students lose points

Check yourself

  1. Question 1

    A student is selected at random from a high school with grades 9 through 12. Which pair of events must be mutually exclusive?

    1. A

      The student takes Spanish; the student takes French.

    2. B

      The student is in grade 9; the student is in grade 12.

    3. C

      The student is in grade 12; the student is on the swim team.

    4. D

      The student takes Spanish; the student is in grade 9.

    Show answer

    B

    Each student is in exactly one grade, so no student is in both grade 9 and grade 12, and the joint probability is 0. Choice A is tempting because the languages are separate classes, but a student can take two languages, so the events can overlap.

  2. Question 2

    For two events in the same random process, P(A)=0.35P(A) = 0.35 and P(B)=0.70P(B) = 0.70. Which statement must be true?

    1. A

      AA and BB are mutually exclusive.

    2. B

      P(A∩B)=0.245P(A \cap B) = 0.245

    3. C

      AA and BB cannot be mutually exclusive.

    4. D

      P(A∩B)=0P(A \cap B) = 0

    Show answer

    C

    If they were disjoint, they would need 0.35+0.70=1.050.35 + 0.70 = 1.05 of the probability, which is more than 1, so they must overlap. Choice B multiplies the probabilities, which is correct only for independent events, and nothing says these are.

  3. Question 3

    A school recorded which of its 240 juniors and seniors have a parking permit: 24 of 120 juniors and 66 of 120 seniors do. One of these 240 students is chosen at random. Find P(senior∩no permit)P(\text{senior} \cap \text{no permit}), and say whether "junior" and "has a permit" are mutually exclusive.

    Show answer

    P(senior∩no permit)=54240=0.225P(\text{senior} \cap \text{no permit}) = \frac{54}{240} = 0.225. "Junior" and "has a permit" are not mutually exclusive, because P(junior∩permit)=24240=0.1>0P(\text{junior} \cap \text{permit}) = \frac{24}{240} = 0.1 > 0.

    Seniors without a permit number 120−66=54120 - 66 = 54, and the selection is from all 240 students, so divide by 240, not by 120. Any positive joint probability rules out mutually exclusive.

  4. Question 4

    Two fair dice are rolled. Are the events "the sum is 7" and "the dice show the same number" mutually exclusive? Justify with a probability.

    Show answer

    Yes. P(sum is 7∩doubles)=0P(\text{sum is 7} \cap \text{doubles}) = 0, so they are mutually exclusive.

    Doubles have sums 2,4,6,8,10,122, 4, 6, 8, 10, 12, all even, and 7 is odd, so none of the 36 outcomes is in both events. The joint probability is 036=0\frac{0}{36} = 0.

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