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12.4 Probability of Compound Events

12.4 Probability of Compound Events. P. 724. Mutually Exclusive Events. Intersection of A & B. To find P(A or B) you must consider what outcomes, if any, are in the intersection of A and B. If there are none, then A and B are mutually exclusive events and P(A or B) = P(A)+P(B)

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12.4 Probability of Compound Events

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  1. 12.4 Probability of Compound Events P. 724

  2. Mutually Exclusive Events

  3. Intersection of A & B

  4. To find P(A or B) you must consider what outcomes, if any, are in the intersection of A and B. • If there are none, then A and B are mutually exclusive events and P(A or B) = P(A)+P(B) • If A and B are not mutually exclusive, then the outcomes in the intersection or A & B are counted twice when P(A) & P(B) are added. • So P(A & B) must be subtracted once from the sum

  5. EXAMPLE 1 • One six-sided die is rolled. • What is the probability of rolling a multiple of 3 or 5? • P(A or B) = P(A) + P(B) = 2/6 + 1/6 = 1/2 • 0.5

  6. EXAMPLE 2 • One six-sided die is rolled. What is the probability of rolling a multiple of 3 or a multiple of 2? • A = Mult 3 = 2 outcomes • B = mult 2 = 3 outcomes • P(A or B) = P(A) + P(B) – P(A&B) • P(A or B) = 2/6 + 3/6 – 1/6 = • 2/3 ≈ 0.67

  7. EXAMPLE 3 • In a poll of high school juniors, 6 out of 15 took a French class and 11 out of 15 took a math class. • Fourteen out of 15 students took French or math. • What is the probability that a student took both French and math?

  8. A = French • B = Math • P(A) = 6/15, P(B) = 11/15, P(AorB) = 14/15 • P(A or B) = P(A) + P(B) – P(A&B) • 14/15 = 6/15 + 11/15 – P(A & B) • P(A & B) = 6/15 + 11/15 – 14/15 • P(A & B) = 3/15 = 1/5 = .20

  9. Using complements to find Probability • The event A’, called the complement of event A, consists of all outcomes that are not in A. • The notation A’ is read ‘A prime’.

  10. Probability of the complement of an event • The probability of the complement of A is : • P(A’) = 1 - P(A)

  11. EXAMPLE 4 • A card is randomly selected from a standard deck of 52 cards. • Find the probability of the given event. • a. The card is not a king. • P(K) = 4/52 so P(K’) = • 48/52 ≈ 0.923

  12. b. The card is not an ace or a jack. • P(not ace or Jack) = 1-(P(4/52 + 4/52)) • = 1- 8/52 = • 44/52 ≈ 0.846

  13. In a survey of 200 pet owners, 103 owned dogs, 88 owned cats, 25 owned birds, and 18 owned reptiles. • 1. None of the respondents owned both a cat and a bird. • What is the probability that they owned a cat or a bird? • 113/200 • = 0.565 • 2. Of the respondents, 52 owned both a cat and a dog. • What is the probability that a respondent owned a cat or a dog? • 139/200 • = 0.695

  14. 3. Of the respondents, 119 owned a dog or a reptile. • What is the probability that they owned a dog and a reptile? • 1/100 = 0.010

  15. Assignment

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