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Independent and Mutually Exclusive Events (modified R. Bloom)

Module by: Roberta Bloom. E-mail the author

Based on: Probability Topics: Independent & Mutually Exclusive Events by Susan Dean, Barbara Illowsky, Ph.D.

Summary: This module explains the concept of independent events, where the probability of event A does not have any effect on the probability of event B, and mutually exclusive events, where events A and B cannot occur at the same time. It is based on the module Probability Topics: Independent and Mutually Exclusive Events from the textbook collection Collaborative Statistics by Dr. Barbara Illowsky and Susan Dean; an example has been added illustrating the determination that two events are not independent.

Independent Events

Two events are independent if the following are true:

  • P(A|B) = P(A)P(A|B) = P(A)
  • P(B|A) = P(B)P(B|A) = P(B)
  • P(A AND B) = P(A) ⋅ P(B)P(A AND B) = P(A) ⋅ P(B)

If events AA and BB are independent, then the chance of AA occurring does not affect the chance of BB occurring and vice versa.

Translating the symbols into words, the first two mathematical statements listed above say that the probability for the event with the condition is the same as the probability for the event without the condition. For independent events: the condition does not change the probability for the event.

For example, two roles of a fair die are independent events. The outcome of the first roll does not change the probability for the outcome of the second roll.

If you select 2 cards consecutively from a complete deck of playing cards, the two selections are not independent; the result of the first selection changes the remaining deck and affects the probabilities for the second selection. This is referred to as selecting "without replacement"; the first card has not been replaced into the deck before the second card is selected.

However, suppose you were to select 2 cards "with replacement", by returning your first card to the deck and shuffling the deck before selecting the second card. Because the deck of cards is complete for both selections, the first selection does not affect the probability of the second selection. When selecting cards with replacement, the selections are independent.

To show that two events are independent, you must show only one of the conditions listed above. If any one of these conditions is true, then all of them are true.

Mutually Exclusive Events

Events AA and BB are mutually exclusive events if they cannot occur at the same time. This means that AA and BB do not share any outcomes and P(A AND B) = 0P(A AND B)= 0.

  • For example, suppose the sample space S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.
  • Let A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7, 8}A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7, 8}, and C = {7, 9}C = {7, 9}.
  • A AND B = {4, 5}A AND B ={4, 5}. P(A AND B) = P(A AND B) = 2 10 2 10 and is not equal to zero. Therefore, AA and BB are not mutually exclusive.
  • AA and CC do not have any numbers in common so P(A AND C) = 0P(A AND C) = 0. Therefore, AA and CC are mutually exclusive.

Note:

Independent and mutually exclusive do not mean the same thing.

You must show that any two events are independent or mutually exclusive. You cannot assume either of these conditions.

If it is not known whether AA and BB are independent or dependent, assume they are dependent until you can show otherwise.

The following examples illustrate these definitions and terms.

Example 1

Flip two fair coins. (This is an experiment.)

The sample space is {HH, HT, TH, TT}{HH, HT, TH, TT} where TT = tails and HH = heads. The outcomes are HHHH, HTHT, THTH, and TTTT. The outcomes HTHT and THTH are different. The HTHT means that the first coin showed heads and the second coin showed tails. The THTH means that the first coin showed tails and the second coin showed heads.

  • Let AA = the event of getting at most one tail. (At most one tail means 0 or 1 tail.) Then AA can be written as {HH, HT, TH}{HH, HT, TH}. The outcome HHHH shows 0 tails. HTHT and THTH each show 1 tail.
  • Let BB = the event of getting all tails. BB can be written as {TT}{TT}. BB is the complement of AA. So, B = A'B = A'. Also, P(A) + P(B) = P(A) + P(A') = 1P(A) + P(B) = P(A) + P(A') = 1.
  • The probabilities for AA and for BB are P(A) = 3 4 P(A) = 3 4 and P(B) = 1 4 P(B)= 1 4 .
  • Let CC = the event of getting all heads. C = {HH}C = {HH}. Since B = {TT}B = {TT}, P(B AND C) = 0P(B AND C) = 0. BB and CC are mutually exclusive. (BB and CC have no members in common because you cannot have all tails and all heads at the same time.)
  • Let DD = event of getting more than one tail. D = {TT}D ={TT}. P(D) = 1 4 P(D)= 1 4 .
  • Let EE = event of getting a head on the first roll. (This implies you can get either a head or tail on the second roll.) E = {HT, HH}E={HT, HH}. P(E) = 2 4 P(E) = 2 4 .
  • Find the probability of getting at least one (1 or 2) tail in two flips. Let FF = event of getting at least one tail in two flips. F= {HT, TH, TT}F={HT, TH, TT}. P(F) = 3 4 P(F) = 3 4

Example 2

Roll one fair 6-sided die. The sample space is {1, 2, 3, 4, 5, 6}{1, 2, 3, 4, 5, 6}. Let event AA = a face is odd. Then A = {1, 3, 5}A = {1, 3, 5}. Let event BB = a face is even. Then B = {2, 4, 6}B = {2, 4, 6}.

  • Find the complement of AA, A'A'. The complement of AA, A'A', is BB because AA and BB together make up the sample space. P(A) + P(B) = P(A) + P(A') = 1P(A) + P(B) = P(A) + P(A') = 1. Also, P(A) = 3 6 P(A)= 3 6 and P(B) = 3 6 P(B)= 3 6
  • Let event CC = odd faces larger than 2. Then C = {3, 5}C={3,5}. Let event DD = all even faces smaller than 5. Then D = {2, 4}D ={2,4}. P(C and D) = 0P(C and D) =0 because you cannot have an odd and even face at the same time. Therefore, CC and DD are mutually exclusive events.
  • Let event EE = all faces less than 5. E = {1, 2, 3, 4}E ={1,2,3,4}.

    Problem 1

    Are CC and EE mutually exclusive events? (Answer yes or no.) Why or why not?

    Solution

    No. CC = {3, 5}{3, 5} and EE = {1, 2, 3, 4}{1, 2, 3, 4}. P(C AND E)=16P(C AND E)=16. To be mutually exclusive, P(C AND E)P(C AND E) must be 0.

  • Find P(C|A)Find P(C|A). This is a conditional. Recall that the event CC is {3, 5}{3, 5} and event AA is {1, 3, 5}{1, 3, 5}. To find P(C|A)P(C|A), find the probability of CC using the sample space AA. You have reduced the sample space from the original sample space {1, 2, 3, 4, 5, 6}{1, 2, 3, 4, 5, 6} to {1, 3, 5}{1, 3, 5}. So, P(C|A) = 2 3 P(C|A) = 2 3

Example 3

Let event GG = taking a math class. Let event HH = taking a science class. Then, G AND HG AND H = taking a math class and a science class. Suppose P(G) = 0.6P(G) =0.6, P(H)= 0.5P(H)= 0.5, and P(G AND H) = 0.3P(G AND H) = 0.3. Are GG and HH independent?

If GG and HH are independent, then you must show ONE of the following:

  • P(G|H) = P(G)P(G|H) = P(G)
  • P(H|G) = P(H)P(H|G) = P(H)
  • P(G AND H)=P(G AND H)= P(G)P(H)P(G)P(H)

Note:

The choice you make depends on the information you have. You could choose any of the methods here because you have the necessary information.

Problem 1

Show that P(G|H) = P(G)P(G|H) = P(G).

Solution

P(G|H) = P(G AND H) P(H) = 0.3 0.5 = 0.6 = P(G) P(G|H) = P(G AND H) P(H) = 0.3 0.5 = 0.6 = P(G)

Problem 2

Show P(G AND H)=P(G AND H)= P(G)P(H)P(G)P(H).

Solution

P(G) P(H)  =  0.6 0.5  =  0.3  =  P(G AND H) P(G)P(H) = 0.60.5 = 0.3 = P(G AND H)

Interpretation of results

Since GG and HH are independent, then, knowing that a person is taking a science class does not change the chance that he/she is taking math. (Note: IF the two events had not been independent - that is, IF they were dependent - then knowing that a person is taking a science class would change the chance he/she is taking math. The next example will illustrate two events that are not independent.)

Example 4

In a particular college class, 60% of the students are female. 50 % of all students in the class have long hair. 45% of the students are female and have long hair. Of the female students, 75% have long hair. Let F be the event that the student is female. Let L be the event that the student has long hair. Are the events of being female and having long hair independent?

  • The following probabilities are given in this example:
  • P(F ) = 0.60P(F ) = 0.60 ; P(L ) = 0.50P(L ) = 0.50
  • P(F AND L) = 0.45P(F AND L) = 0.45
  • P(L|F) = 0.75P(L|F) = 0.75

If FF and LL are independent, then the following conditions are true. BUT if you show that any of the conditions are not true, then FF and LL are not independent.

  • P(L|F) = P(L)P(L|F) = P(L)
  • P(F|L) = P(F)P(F|L) = P(F)
  • P(F AND L)=P(F AND L)= P(F)P(L)P(F)P(L)

Note:

The choice you make depends on the information you have. You could use the first or last condition on the list for this example. You do not know P(F|L) yet, so you can not use the second condition.

Solution 1

Check whether P(F and L) = P(F)P(L): We are given that P(F and L) = 0.45 ; but P(F)P(L) = (0.60)(0.50)= 0.30 The events of being female and having long hair are not independent because P(F and L) does not equal P(F)P(L).

Solution 2

Check whether P(L|F) equals P(L): We are given that P(L|F) = 0.75 but P(L) = 0.50; they are not equal. The events of being female and having long hair are not independent.

Interpretation of results

The events of being female and having long hair are not independent; knowing that a student is female changes the probability that a student has long hair.

Example 5

In a box there are 3 red cards and 5 blue cards. The red cards are marked with the numbers 1, 2, and 3, and the blue cards are marked with the numbers 1, 2, 3, 4, and 5. The cards are well-shuffled. You reach into the box (you cannot see into it) and draw one card.

Let RR = red card is drawn, BB = blue card is drawn, EE = even-numbered card is drawn.

The sample space S  =  R1, R2, R3, B1, B2, B3, B4, B5 S = R1, R2, R3, B1, B2, B3, B4, B5. SS has 8 outcomes.

  • P(R) = 3 8 P(R)= 3 8 . P(B) = 5 8 P(B)= 5 8 . P(R AND B) = 0 P(R AND B)=0. (You cannot draw one card that is both red and blue.)
  • P(E) = 3 8 P(E)= 3 8 . (There are 3 even-numbered cards, R2R2, B2B2, and B4B4.)
  • P(E|B) = 2 5 P(E|B)= 2 5 . (There are 5 blue cards: B1B1, B2B2, B3B3, B4B4, and B5B5. Out of the blue cards, there are 2 even cards: B2B2 and B4B4.)
  • P(B|E) = 2 3 P(B|E)= 2 3 . (There are 3 even-numbered cards: R2R2, B2B2, and B4B4. Out of the even-numbered cards, 2 are blue: B2B2 and B4B4.)
  • The events RR and BB are mutually exclusive because P(R AND B) = 0P(R AND B) = 0.
  • Let GG = card with a number greater than 3. G = {B4, B5}G ={B4,B5}. P(G) = 2 8 P(G)= 2 8 . Let HH = blue card numbered between 1 and 4, inclusive. H = {B1,B2,B3, B4}H ={B1,B2,B3,B4}. P(G|H) = 1 4 P(G|H)= 1 4 . (The only card in H that has a number greater than 3 is B4B4.) Since 2 8 = 1 4 2 8 = 1 4 , P(G) = P(G|H)P(G) =P(G|H) which means that GG and HH are independent.

Glossary

Independent Events:
The occurrence of one event has no effect on the probability of the occurrence of any other event. Events A and B are independent if one of the following is true: (1). P ( A | B ) = P ( A ) ;P( A | B)=P(A); (2) P ( B | A ) = P ( B ) ;P( B | A)=P(B); (3) P ( A and B ) = P ( A ) P ( B )P(AandB)=P(A)P(B).
Mutually Exclusive:
An observation cannot fall into more than one class (category). Being in one category prevents being in a mutually exclusive category.

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