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Lesson 5 - 2

Lesson 5 - 2. Addition Rule and Complements. Objectives. Use the Addition Rule for disjoint events Use the General Addition Rule Compute the probability of an event using the Complement Rule. Vocabulary.

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Lesson 5 - 2

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  1. Lesson 5 - 2 Addition Rule and Complements

  2. Objectives • Use the Addition Rule for disjoint events • Use the General Addition Rule • Compute the probability of an event using the Complement Rule

  3. Vocabulary • Disjoint – mutually exclusive events; two events are disjoint if they have no outcomes in common • Benford’s Law – assigned probabilities to the digits (1-9) occurring first in a number • Contingency table – relates two categories of data (one in the rows and the other between two columns) • Complement of an event – all possible outcomes not in that event

  4. Addition Rule for Disjoint Events If E and F are disjoint (mutually exclusive) events, then P(E or F) = P(E) + P(F) Probability for Disjoint Events E F P(E or F) = P(E) + P(F)

  5. General Addition Rule For any two events E and F, P(E or F) = P(E) + P(F) – P(E and F) F E E and F Probability for non-Disjoint Events P(E or F) = P(E) + P(F) – P(E and F)

  6. Complement Rule If E represents any event and Ec represents the complement of E, then P(Ec) = 1 – P (E) Probability for Complement Events E Ec P(Ec) = 1 – P(E)

  7. Example 1 A card is chosen at random from a normal deck. What is the probability of choosing? a) a king or a queen b) a king and a queen c) a king and red card d) a king or red card d) a face card and a heart e) a face card or a heart P(K) + P(Q) = 4/52 + 4/52 = 8/52 ≈ 15.4% P(K+Q) = 0 P(K+red) = (4/52)•(26/52) = 2/52 ≈ 3.8% P(K)+P(red) = 4/52 + 26/52 - 2/52 = 28/52 ≈ 53.8% P(K,Q,J)+P(red) = 12/52 + 13/52 – 3/52 = 22/52 ≈ 42.3% P(K,Q,J + heart) = (12/52) •(13/52) = 3/52 ≈ 5.8%

  8. Example 2 Fifty animals are to be used in a stress study: 4 male and 6 female dogs, 9 male and 7 female cats, 5 male and 8 female monkeys, 6 male and 5 female rats. Find the probability of choosing: a) a dog or a cat b) a cat or a female c) a male d) a monkeys or a male P(D) + P(C) = 10/50 + 16/50 = 26/50 = 52% P(C) + P(F) = P(C) + P(F) – P(C & F) = 10/50 + 26/50 – 7/50 = 29/50 = 58% P(Male) = 24/50 = 48% P(M)+P(male) = P(M) + P(m) – P(M&m) = 13/50 + 24/50 - 5/50 = 32/50 = 64%

  9. Example 2 cont Fifty animals are to be used in a stress study: 4 male and 6 female dogs, 9 male and 7 female cats, 5 male and 8 female monkeys, 6 male and 5 female rats. Find the probability of choosing: e) an animal other than a female monkey f) a female or a rat g) a female and a cat h) a dog and a cat 1 – P(f&M) = 1 – 8/50 = 42/50 = 84% P(f) + P(R) = P(f) + P(R) – P(f & R) = 26/50 + 11/50 – 5/50 = 32/50 = 64% P(female&C) = (7/16)•(16/50) = 7/50 = 14% P(D&C) = 0%

  10. Example 3 Customers at a certain department store pay for purchases either with cash or one of four types of credit cards. See table below. Mode of payment : Cash Store card MC V AE Probability .30 .25 .18 .15 .12 a) Let N be the event that the next purchase is made with a national credit card. What is P(N)? b) Let S be the event that the purchase is with the store card. What is P( not S)? P(N) = P(MC) + P(V) + P(AE) = 0.18 + 0.15 + 0.12 = 45% P(~S) = 1 - P(S) = 1 – 0.25 = 75%

  11. Example 4 A pollster surveys 100 subjects consisting of 40 Dems (of which half are female) and 60 Reps (half are female). What is the probability of randomly selecting one of these subjects of getting: a) a Dem b) a female c) a Dem and a female d) a Rep male e) a Dem or a male e) a Rep or a female P(f) = P(f&D) + P(f&R) = 20/100 + 30/100 = 50% P(D) = 40/100 = 40% P(D&f) = 0.4 * 0.5 or 20/100 = 20% P(R&m) = 0.6 * 0.5 or 30/100 = 30% P(D) + P(m) = = 40/100 + 50/100 – 20/100 = 70% P(R)+P(f) = 60/100 + 50/100 – 30/100 = 80/100 = 80%

  12. Example 5 Suppose that 60% of all customers of a large insurance agency have automobile policies with the agency, 40% have homeowners policies and 25% have both. If a customer is selected at random, what is the probability that the customer has: a) has both policies b) has only a homeowners policy c) has only an auto policy d) has no policy at all 25% P(h&a) = 0.25 40 – 40/4 = 30% P(h) = 0.30 P(a) = 60 – 60/4 = 45% P(a) = 0.45 P(~p) = 0%

  13. Summary and Homework • Summary • Probabilities obey additional rules • For disjoint events, the Addition Rule is used for calculating “or” probabilities • For events that are not disjoint, the Addition Rule is not valid … instead the General Addition Rule is used for calculating “or” probabilities • The Complement Rule is used for calculating “not” probabilities • Homework • pg 274-277: 5, 8, 9, 14, 17, 20, 27, 30, 36, 40, 41

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