1 / 15

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)

maj
Télécharger la présentation

12.4 Probability of Compound Events

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related