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Example: A box contains 12 colored balls, of which 5 are white, 4 are black, and 3 are red. Sample 7 balls with replacement. • n = 7 trials, k= 3 possible outcomes E₁= white ball, with probability p₁ = 5/12 = black ball, with probability p₂ = 4/12 • E3 = red ball, with probability p3 = 3/12 E₂ = ● 5 4 3 √ (31-32-33-7- 12.12.12) - ( X2, 7, where x₁ + x₂ + x3 = 7 7 5 (6) X1, X2, X312 12 12 O 16

Example: A box contains 12 colored balls, of which 5 are white, 4 are black, and 3 are red. Sample 7 balls with replacement. • n = 7 trials, k= 3 possible outcomes E₁= white ball, with probability p₁ = 5/12 = black ball, with probability p₂ = 4/12 • E3 = red ball, with probability p3 = 3/12 E₂ = ● 5 4 3 √ (31-32-33-7- 12.12.12) - ( X2, 7, where x₁ + x₂ + x3 = 7 7 5 (6) X1, X2, X312 12 12 O 16

College Algebra
College Algebra
7th Edition
ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter9: Counting And Probability
Section9.3: Binomial Probability
Problem 2E: If a binomial experiment has probability p success, then the probability of failure is...
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Sample 7 balls with replacement
Example: A box contains 12 colored balls, of which 5 are white, 4 are black, and 3 are red. Sample 7 balls with replacement. • n = 7 trials, k= 3 possible outcomes E₁= white ball, with probability p₁ = 5/12 = black ball, with probability p₂ = 4/12 • E3= red ball, with probability p3= 3/12 E₂ = ● 5 4 3 12 12 12 where x₁ + x₂ + x3 = 7 √51,372,48357, 1, X2, X3; 7, 7 5 L666 X1, X2, X312 12 O 12 16
Transcribed Image Text:Example: A box contains 12 colored balls, of which 5 are white, 4 are black, and 3 are red. Sample 7 balls with replacement. • n = 7 trials, k= 3 possible outcomes E₁= white ball, with probability p₁ = 5/12 = black ball, with probability p₂ = 4/12 • E3= red ball, with probability p3= 3/12 E₂ = ● 5 4 3 12 12 12 where x₁ + x₂ + x3 = 7 √51,372,48357, 1, X2, X3; 7, 7 5 L666 X1, X2, X312 12 O 12 16
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