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M ARIO F . T RIOLA

S TATISTICS. E LEMENTARY. Section 4-3 Binomial Probability Distributions. M ARIO F . T RIOLA. E IGHTH. E DITION. Binomial Probability Distribution 1. The experiment must have a fixed number of trials .

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M ARIO F . T RIOLA

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  1. STATISTICS ELEMENTARY Section 4-3 Binomial Probability Distributions MARIO F. TRIOLA EIGHTH EDITION

  2. Binomial Probability Distribution 1. The experiment must have a fixed number of trials. 2. The trials must be independent. (The outcome of any individual trial doesn’t affect the probabilities in the other trials.) 3. Each trial must have all outcomes classified into two categories. 4. The probabilities must remain constantfor each trial. Definitions

  3. n = fixed number of trials x = specific number of successes in n trials p = probability of success in one of n trials q = probability of failure in one of n trials (q = 1 - p) P(x) = probability of getting exactly x success among n trials Be sure that x and p both refer to the same category being called a success. Notation for Binomial Probability Distributions

  4. Method 1 Binomial Probability Formula n! • P(x) = • px• qn-x (n - x )! x!

  5. Method 1 Binomial Probability Formula n! • P(x) = • px• qn-x (n - x )! x! • P(x) = nCx• px•qn-x for calculators with nCr key, where r = x

  6. This is a binomial experiment where: n = 5 x = 3 p = 0.90 q = 0.10 Example: Find the probability of getting exactly 3 correct responses among 5 different requests from AT&T directory assistance. Assume in general, AT&T is correct 90% of the time.

  7. This is a binomial experiment where: n = 5 x = 3 p = 0.90 q = 0.10 Using the binomial probability formula to solve: P(3) = 5C3• 0.9 • 01 = 0.0.0729 Example: Find the probability of getting exactly 3 correct responses among 5 different requests from AT&T directory assistance. Assume in general, AT&T is correct 90% of the time. 2 3

  8. For n = 15 and p = 0.10 Table A-1 P(x) n x 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0.206 0.343 0.267 0.129 0.043 0.010 0.002 0.0+ 0.0+ 0.0+ 0.0+ 0.0+ 0.0+ 0.0+ 0.0+ 0.0+ 15

  9. For n = 15 and p = 0.10 Table A-1 P(x) n x 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0.206 0.343 0.267 0.129 0.043 0.010 0.002 0.0+ 0.0+ 0.0+ 0.0+ 0.0+ 0.0+ 0.0+ 0.0+ 0.0+ 15

  10. For n = 15 and p = 0.10 Table A-1 Binomial Probability Distribution P(x) x P(x) n x 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0.206 0.343 0.267 0.129 0.043 0.010 0.002 0.0+ 0.0+ 0.0+ 0.0+ 0.0+ 0.0+ 0.0+ 0.0+ 0.0+ 0.206 0.343 0.267 0.129 0.043 0.010 0.002 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 15

  11. a) P(3) = 0.073 b) P(at least 3) = P(3 or 4 or 5) = P(3) or P(4) or P(5) = 0.073 + 0.328 + 0.590 = 0.991 Example: Using Table A-1 for n = 5 and p = 0.90, find the following: a) The probability of exactly 3 successesb) The probability of at least 3 successes

  12. Binomial Probability Formula n! P(x) = • px•qn-x (n - x )! x! Number of outcomes with exactly x successes among n trials

  13. Binomial Probability Formula n! P(x) = • px•qn-x (n - x )! x! Probability of x successes among n trials for any one particular order Number of outcomes with exactly x successes among n trials

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