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Chapter 5: Discrete Probability Distributions

Chapter 5: Discrete Probability Distributions. Section 5.4: The Binomial Distribution. A binomial experiment is a probability distribution That satisfies the following four requirements:. two outcomes. Each trial can have only __________________ or

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Chapter 5: Discrete Probability Distributions

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  1. Chapter 5: Discrete Probability Distributions Section 5.4: 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 :probability of success q :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: P(X) = TotalC# you want ·(prob of success)(# of successes)·(prob of failure)(# failures)

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

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

  8. Round to the nearest thousandth EXAMPLE 2: If a student randomly guesses at six multiple-choicequestions, find the probability that the student getsexactly three correct. Each question has 5 choices. P(3 correct) = nCX · pX · q(n – X) success = correct n = 6 X (# of successes)= 3 1 – (1/5) = 4/5 p = 1/5 q = P(3 correct) = 6C3 · (1/5)3 · (4/5)(6 – 3) 20 · (0.008) · (0.512) = 0.082

  9. 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 0.3 1 – 0.7= p = 0.7 q = P(5) = 7C5 · (0.7)5 · (0.3)(7 – 5) · (0.16807) · (0.09) 21 = 0.318

  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) = 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

  11. 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

  12. p247 # 1 -See if each of the parts qualify as a binomial distribution (4 parts from our notes)

  13. p247 # 1, 5, 7, 9

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