1 / 10

Chapter 5: Discrete Probability Distributions

Chapter 5: Discrete Probability Distributions. Section 5.3: The Binomial Distribution. A binomial experiment is a probability distribution That satisfies the following four requirements:. two outcomes. Each trial can have only __________________ or

anakin
Download Presentation

Chapter 5: Discrete Probability Distributions

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. Chapter 5: Discrete Probability Distributions Section 5.3: The Binomial Distribution

  2. A binomial experiment is a probability distribution That satisfies the following four requirements: two outcomes • Each trial can have only __________________ or • outcomes that can be reduced to two outcomes.These outcomes can be considered as either _______________ or _____________. success failure

  3. A binomial experiment is a probability distribution That satisfies the following four requirements: fixed number 2) There must be a _______________ of trials. independent • The outcomes of each trial must be ____________ • of each other remain the same 4) The probability of a success must ________________for each trial.

  4. Symbols for the Binomial Distribution p :Numerical probability of success q :Numerical probability of failure n :number of trials X :number of successes *Note: q = 1 – p

  5. Binomial Probability Formula In a binomial experiment, the probability of exactly X successes in n trials is: ___n!__(n-X)! X! P(X) = · pX · q(n – X) or P(X) = nCX · pX · q(n – X)

  6. Round to the nearest thousandth EXAMPLE 1: A coin is tossed three times. Find the probabilityof getting exactly one tail. P(1 tail) = nCX · pX · q(n – X) success = tail n = 3 X (# of successes)= 1 p = 0.5 q = 0.5 P(1 tail) = 3C1 · (.5)1 · (.5)(3 – 1) 3 · 0.5 · 0.25 = 0.375

  7. Round to the nearest thousandth EXAMPLE 2: If a student randomly guesses at five multiple-choicequestions, find the probability that the student getsexactly three correct. Each questions has 5 choices. P(3 correct) = nCX · pX · q(n – X) success = correct n = 5 X (# of successes)= 3 p = 1/5 q = 4/5 P(3 correct) = 5C3 · (1/5)3 · (4/5)(5 – 3) 10 · (0.008) · (0.64) = 0.051

  8. Round to the nearest thousandth EXAMPLE 3: A surgical technique is performed on seven patients.You are told there is a 70% chance of success. Find the probability that the surgery is successful for: a. Exactly five patients P(5) = nCX · pX · q(n – X) success = Surgery success n = 7 X (# of successes)= 5 p = 0.7 q = 0.3 P(5) = 7C5 · (0.7)5 · (0.3)(7 – 5) 21 · (0.16807) · (0.09) = 0.318

  9. EXAMPLE 3: A surgical technique is performed on seven patients.You are told there is a 70% chance of success. Find the probability that the surgery is successful for: b. At least five patients P(5) = 7C5 · (0.7)5 · (0.3)(7 – 5) nCX · pX · q(n – X) = 21 · (0.16807) · (0.09) = 0.318 P(6) = 7C6 · (0.7)6 · (0.3)(7 – 6) nCX · pX · q(n – X) = 7 · (0.11765) · (0.3) = 0.247 P(7) = 7C7 · (0.7)7 · (0.3)(7 – 7) nCX · pX · q(n – X) = 1 · (0.08235) · (1) = 0.082

  10. EXAMPLE 3: A surgical technique is performed on seven patients.You are told there is a 70% chance of success. Find the probability that the surgery is successful for: b. At least five patients P(5) + P(6) + P(7) = 0.318 + 0.247 + 0.082 = 0.647

More Related