6.5 conditional probability and independence

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6.5 Conditional Probability and Independence 1. Conditional Probability 2. Conditional Probability of Equally Likely Outcomes 3. Product Rule 4. Independence 5. Independence of a Set of Events 1

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6.5 Conditional Probability and Independence. Conditional Probability Conditional Probability of Equally Likely Outcomes Product Rule Independence Independence of a Set of Events. Conditional Probability. - PowerPoint PPT Presentation

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Page 1: 6.5  Conditional Probability and Independence

6.5 Conditional Probability and Independence

1. Conditional Probability2. Conditional Probability of Equally Likely

Outcomes 3. Product Rule4. Independence5. Independence of a Set of Events

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Page 2: 6.5  Conditional Probability and Independence

Conditional ProbabilityLet E and F be events is a sample space S. The conditional probability Pr(E | F) is the probability of event E occurring given the condition that event F has occurred. In calculating this probability, the sample space is restricted to F.

Pr( )Pr( | )Pr( )E FE FF

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provided that Pr(F) ≠ 0.

Page 3: 6.5  Conditional Probability and Independence

Example Conditional Probability

Twenty percent of the employees of Acme Steel Company are college graduates. Of all its employees, 25% earn more than $50,000 per year, and 15% are college graduates earning more than $50,000. What is the probability that an employee selected at random earns more than $50,000 per year, given that he or she is a college graduate?

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Example Conditional Probability (2)

Let H = "earns more than $50,000 per year" and C = "college graduate."From the problem,Pr(H) = .25 Pr(C) = .20 Pr(H C) = .15.Therefore, Pr( ) .15 3Pr( | ) .

Pr( ) .20 4H CH CC

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Page 5: 6.5  Conditional Probability and Independence

Conditional Probability - Equally Likely Outcomes

Conditional Probability in Case of Equally Likely Outcomes

number of outcomes in Pr( | )

number of outcomes in E F

E FF

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provided that [number of outcomes in F] ≠ 0.

Page 6: 6.5  Conditional Probability and Independence

Example Conditional Probability

A sample of two balls are selected from an urn containing 8 white balls and 2 green balls. What is the probability that the second ball selected is white given that the first ball selected was white?

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Page 7: 6.5  Conditional Probability and Independence

Example Conditional Probability

The number of outcomes in "the first ball is white" is 89 = 72. That is, the first ball must be among the 8 white balls and the second ball can be any of the 9 balls left.The number of outcomes in "the first ball is white and the second ball is white" is 87 = 56.

56 7Pr(2 ball is white |1 ball is white)72 9

nd st

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Page 8: 6.5  Conditional Probability and Independence

Product RuleProduct Rule If Pr(F) ≠ 0,

Pr(E F) = Pr(F) Pr(E | F).

The product rule can be extended to three events.Pr(E1 E2 E3) = Pr(E1) Pr(E2 | E1) Pr(E3| E1 E2)

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Page 9: 6.5  Conditional Probability and Independence

Example Product RuleA sequence of two playing cards is drawn at random (without replacement) from a standard deck of 52 cards. What is the probability that the first card is red and the second is black?LetF = "the first card is red," andE = "the second card is black."

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Page 10: 6.5  Conditional Probability and Independence

Example Product Rule (2)

Pr(F) = ½ since half the deck is red cards.

26Pr( | ) .51

E F

1 26 13Pr( ) .2 51 51

E F

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If we know the first card is red, then the probability the second is black is

Therefore,

Page 11: 6.5  Conditional Probability and Independence

Independence Let E and F be events. We say that E and F are independent provided that

Pr(E F) = Pr(E) Pr(F).Equivalently, they are independent provided that

Pr(E | F) = Pr(E) and Pr(F | E) = Pr(F).

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Page 12: 6.5  Conditional Probability and Independence

Example IndependenceLet an experiment consist of observing the results of drawing two consecutive cards from a 52-card deck. Let E = "second card is black" and F = "first card is red". Are these two events independent?From the previous example, Pr(E | F) = 26/51.Note that Pr(E) = 1/2. Since they are not equal, E and F are not independent.

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Page 13: 6.5  Conditional Probability and Independence

Independence of a Set of Events

A set of events is said to be independent if, for each collection of events chosen from them, say E1, E2, …, En, we have

Pr(E1 E2 … En) = Pr(E1) Pr(E2) … Pr(En).

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Page 14: 6.5  Conditional Probability and Independence

Example Independence of a Set

A company manufactures stereo components. Experience shows that defects in manufacture are independent of one another. Quality control studies reveal that2% of CD players are defective, 3% of amplifiers are defective, and 7% of speakers are defective.A system consists of a CD player, an amplifier, and 2 speakers. What is the probability that the system is not defective?

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Page 15: 6.5  Conditional Probability and Independence

Example Independence of a Set (2)

Let C, A, S1, and S2 be events corresponding to defective CD player, amplifier, speaker 1, and speaker 2, respectively. ThenPr(C) = .02, Pr(A) = .03, Pr(S1) = Pr(S2) = .07

Pr(C') = .98, Pr(A') = .97, Pr(S1') = Pr(S2') = .93

Pr(C‘ A‘ S1‘ S2') = .98.97.932 = .822

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Page 16: 6.5  Conditional Probability and Independence

Summary Section 6.5 - Part 1

Pr(E|F), the conditional probability that E occurs given that F occurs, is computed as Pr(E F)/Pr(F). For a sample space with a finite number of equally likely outcomes, it can be computed as n(E F)/n(F). The product rule states that if Pr(F) ≠ 0, then Pr(E F) = Pr(F) Pr(E|F).

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Page 17: 6.5  Conditional Probability and Independence

Summary Section 6.5 - Part 2

E and F are independent events if Pr(E F) = Pr(F) Pr(E). Equivalently, E and F [with Pr(F) ≠ 0] are independent events if Pr(E|F) = Pr(E). A collection of events is said to be independent if for each collection of events chosen from them, the probability that all the events occur equals the product of the probabilities that each occurs.

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