Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 10 Counting Methods and Probability - 10.6 Construct and Interpret Binomial Distributions - 10.6 Exercises - Problem Solving - Page 729: 45d

Answer

See below

Work Step by Step

The standard formula: $P(number-of-successes)=\frac{n!}{(n-k)!k!}.p^k(1-p)^{n-k}$ The probability that someone is Rh+ is $p=0.37+0.34+0.1+0.04=0.85$ It means that 5,6,7,8,9,10 can be type Rh+. Substituting $p=0.85,n=10,k=5$ we have: $P(5)=\frac{10!}{(10-5)!5!}.(0.85)^5(1-0.85)^{10-5}\approx0.0085$ Substituting $p=0.5,n=10,k=6$ we have: $P(6)=0.0401$ Substituting $p=0.5,n=10,k=7$ we have: $P(7)=0.1298$ Substituting $p=0.5,n=10,k=8$ we have: $P(8)=0.2759$ Substituting $p=0.5,n=10,k=9$ we have: $P(9)=0.3474$ Substituting $p=0.5,n=10,k=10$ we have: $P(10)=0.1969$ Hence, $P(5,6,7,8,9,10)=0.0085+0.0401+0.1298+0.2759+0.3474+0.1969=0.9986$ The probability that at least 5 people are Rh+ is $0.9986$
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