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Probability: Independent and Mutually Exclusive Events is part of the collection col10555 written by Barbara Illowsky and Susan Dean and 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. The module has contributions from Roberta Bloom.

Independent and mutually exclusive do not mean the same thing.

Independent events

Two events are independent if the following are true:

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

Two events A and B are independent if the knowledge that one occurred does not affect the chance the other occurs. For example, the outcomes of 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 secondroll. To show two events are independent, you must show only one of the above conditions. If two events are NOT independent, then we say that they are dependent .

Sampling may be done with replacement or without replacement .

  • With replacement : If each member of a population is replaced after it is picked, then that member has the possibility of being chosen more than once. When sampling is done with replacement, then events are considered to be independent, meaning the result of the first pick will not change the probabilities for the second pick.
  • Without replacement: : When sampling is done without replacement, then each member of a population may be chosen only once. In this case, the probabilities for the second pick are affected by the result of the first pick. The events are considered to be dependent or not independent.

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

Mutually exclusive events

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

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

If it is not known whether A and B are mutually exclusive, assume they are not until you can show otherwise .

The following examples illustrate these definitions and terms.

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

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

  • Let A = the event of getting at most one tail . (At most one tail means 0 or 1 tail.) Then A can be written as {HH, HT, TH} . The outcome HH shows 0 tails. HT and TH each show 1 tail.
  • Let B = the event of getting all tails. B can be written as {TT} . B is the complement of A . So, B = A' . Also, P(A) + P(B) = P(A) + P(A') = 1 .
  • The probabilities for A and for B are P(A) = 3 4 and P(B) = 1 4 .
  • Let C = the event of getting all heads. C = {HH} . Since B = {TT} , P(B AND C) = 0 . B and C are mutually exclusive. ( B and C have no members in common because you cannot have all tails and all heads at the same time.)
  • Let D = event of getting more than one tail. D = {TT} . P(D) = 1 4 .
  • Let E = 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 } . P(E) = 2 4 .
  • Find the probability of getting at least one (1 or 2) tail in two flips. Let F = event of getting at least one tail in two flips. F = { HT , TH , TT } . P(F) = 3 4
Practice Key Terms 2

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Source:  OpenStax, Engr 2113 ece math. OpenStax CNX. Aug 27, 2010 Download for free at http://cnx.org/content/col11224/1.1
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