Skip to content Skip to navigation

OpenStax-CNX

You are here: Home » Content » Probability Topics: Independent & Mutually Exclusive Events

Navigation

Recently Viewed

This feature requires Javascript to be enabled.

Tags

(What is a tag?)

These tags come from the endorsement, affiliation, and other lenses that include this content.
 

Probability Topics: Independent & Mutually Exclusive Events

Module by: Susan Dean, Barbara Illowsky, Ph.D.. E-mail the authors

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.

Note: You are viewing an old version of this document. The latest version is available here.

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 AA and BB are independent, then the chance of AA occurring does not affect the chance of BB occurring and vice versa. 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. To show two events are independent, you must show only one of the above conditions.

Mutually Exclusive 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)

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. If the two events had not been independent (that is, they are dependent) then knowing that a person is taking a science class would change the chance he/she is taking math. For practice, show that P(H|G) = P(H)P(H|G) = P(H) to show that GG and HH are independent events.

Example 4

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.

Content actions

Download module as:

Add module to:

My Favorites (?)

'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'.

| A lens I own (?)

Definition of a lens

Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

Who can create a lens?

Any individual member, a community, or a respected organization.

What are tags? tag icon

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

| External bookmarks