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 - Skill Practice - Page 728: 32

Answer

See below

Work Step by Step

The standard formula: $P(number-of-successes)=\frac{n!}{(n-k)!k!}.p^k(1-p)^{n-k}$ Substituting $p=0.36,n=10,k=3$ we have: $P(3)=\frac{10!}{(10-3)!3!}.(0.36)^3(1-0.36)^{10-3}\approx0.2426$ Substituting $p=0.36,n=10,k=4$ we have: $P(4)=\frac{10!}{(10-4)!4!}.(0.36)^4(1-0.36)^{10-4}\approx0.24239$ Substituting $p=0.36,n=10,k=6$ we have: $P(6)=\frac{10!}{(10-6)!6!}.(0.36)^6(1-0.36)^{10-6}\approx0.07669$ Substituting $p=0.36,n=10,k=7$ we have: $P(7)=\frac{10!}{(10-7)!7!}.(0.36)^7(1-0.36)^{10-7}\approx0.02465$ We can see that 3 is the most likely to be the number of successes. Hence, the correct option is A.
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