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Binomial Probability Distributions

This section covers the basics of binomial probability distributions, including the properties and formulas used to calculate probabilities. Examples and methods for solving binomial probability problems are provided, along with instructions for using a calculator.

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Binomial Probability Distributions

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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 constant for each trial. Definitions

  3. P(x) = probability of getting exactly x success among n trials 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) 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! • P(x) = nCx• px•qn-x for calculators with nCr function, r = x

  5. Method 1 – Using a formula 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

  6. Method 1 – Using a formula 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 • 0.1 = 0.0729 Method 1 – Using a formula 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 Method 2 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

  9. a) P(3) = 0.129 b) P(at most 3) = P(0 or 1 or 2 or 3) = P(2) or P(1) or P(2) or P(3) = 0.206 + 0.343 + 0.267 + 0.129 = 0.945 Note = This method is limited because a table may not be available for every n and/or p. Method 2 – Using a table Example: Using Table A-1 for n = 15 and p = 0.10, find the following: a) The probability of exactly 3 successesb) The probability of at most 3 successes

  10. Press 2nd,VARS (DISTR). Select the option binompdf(). Complete the entry binompdf(n, p, x) to obtain P(x). n is the number of trials p is the probability of success x is the EXACT number of successes. Method 3 – Using TI-83/4 Probabilities with “Exact” successes

  11. Example: What is the probability of getting exactly 2 heads when 4 tosses are made? Solution: P(2) = binompdf(4, 0.5, 2) P(2) = 0.375 Method 3 - Using TI-83/4 Probabilities with “Exact” successes

  12. Example: What is the probability of getting at most 2 heads when 4 tosses are made? Express at most 2 as an inequality. P( x ≤ 2) which means x = 0 or 1 or 2 Solution: P( x ≤ 2) = P(0) + P(1) + P(2) P( x ≤ 2) = 0.0625 + 0.25 + 0.375 = 0.6875 Where the probabilities would computed using binompdf(4,0.5, 0) then binompdf(4,0.5, 1) etc… Method 3 - Using TI-83/4 Probabilities with “At most” successes

  13. Press 2nd,VARS, select the option binomcdf(). Note: The “c” indicates this is a cumulative function. It adds all the probabilities from zero up to x number of successes. Complete the entry to obtain P(At most x) = binomcdf(n, p, x), where x is the MAXIMUM number of successes. Method 3 - Using TI-83/4 Probabilities with “At most” successes

  14. Example: What is the probability of getting at most 2 heads when 4 tosses are made? Solution: P( x ≤ 2) = binomcdf(4, 0.5, 2) = 0.6875. Method 3 - Using TI-83/4

  15. When doing at least problems we must use the complement rule P(A) = 1 – P(not A) Complete the entry P(At least x) = 1 - binomcdf(n, p, x- 1), where x is the MINIMUM number of successes. Method 3 - Using TI-83/4 Probabilities with “At least” successes

  16. Example: What is the probability of getting at least 3 heads when 4 tosses are made? Solution: P(x≥3) = 1 – P(x ≤ 2) P(x≥3) = 1 - binomcdf(4, 0.5, 2) = 0.3125. Note: This is the same as P( x ≥ 3)= P(x=3)+ P(x=4) P( x ≥ 3)= 0.25 + 0.0625 = 0.3125 Method 3 - Using TI-83/4

  17. Recap • P(x) Ask to find the probability of EXACT number of successes. • Formula: P(x) = nCx·px·qn-x • Calculator: P(x) = binompdf(n,p,x) • P(X  x) Ask to find the probability of AT MOST a number of successes. • Calculator: P(X  x ) = binomcdf(n, p, x) • P(X  x) Ask to find the probability of AT LEAST a number of successes. • Calculator: P(X  x ) = 1 - binomcdf(n, p, x-1)

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