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6.5 Conditional Probability and Independence

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.

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6.5 Conditional Probability and Independence

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  1. 6.5 Conditional Probability and Independence • Conditional Probability • Conditional Probability of Equally Likely Outcomes • Product Rule • Independence • Independence of a Set of Events

  2. Conditional Probability • Let E and F be events is a sample space S. The conditional probabilityPr(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. provided that Pr(F) ≠ 0.

  3. 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?

  4. 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,

  5. Conditional Probability - Equally Likely Outcomes • Conditional Probability in Case of Equally Likely Outcomes provided that [number of outcomes in F] ≠ 0.

  6. 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?

  7. Example Conditional Probability • The number of outcomes in "the first ball is white" is 89 = 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 87 = 56.

  8. Product Rule • Product Rule If Pr(F) ≠ 0, • Pr(EF) = Pr(F)Pr(E|F). • The product rule can be extended to three events. • Pr(E1E2E3) = Pr(E1)Pr(E2|E1)Pr(E3| E1 E2)

  9. Example Product Rule • A 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? • Let • F = "the first card is red," and • E = "the second card is black."

  10. Example Product Rule (2) • Pr(F) = ½ since half the deck is red cards. If we know the first card is red, then the probability the second is black is Therefore,

  11. Independence • Let E and F be events. We say that E and F are independent provided that • Pr(EF) = Pr(E)Pr(F). • Equivalently, they are independent provided that • Pr(E|F) = Pr(E) and Pr(F|E) = Pr(F).

  12. Example Independence • Let 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.

  13. 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(E1E2 … En) = Pr(E1)Pr(E2) … Pr(En).

  14. 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 that • 2% 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?

  15. 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. Then • Pr(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

  16. Summary Section 6.5 - Part 1 • Pr(E|F), the conditional probability that E occurs given that F occurs, is computed as Pr(EF)/Pr(F). For a sample space with a finite number of equally likely outcomes, it can be computed as n(EF)/n(F). • The product rule states that if Pr(F) ≠ 0, then Pr(EF) = Pr(F) Pr(E|F).

  17. Summary Section 6.5 - Part 2 • E and F are independent events if Pr(EF) = 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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