Statistics for the Life Sciences (5th Edition)

Published by Pearson
ISBN 10: 0-32198-958-9
ISBN 13: 978-0-32198-958-1

Chapter 3 - Probability and the Binomial Distribution - Exercises 3.6.1 - 3.6.12: 3.6.4

Answer

P(X=j) = $\frac{n!}{j!(n-j)!}$$p^{j}$$(1-p)^{n-j}$

Work Step by Step

a.) P(all 20 will be cured) = $\frac{20!}{20!0!}$$0.9^{20}$$(1-0.9)^{0}$ = 0.1216 b.) P(all but 1 will be cured) = $\frac{20!}{19!1!}$$0.9^{19}$$(1-0.9)^{1}$ = 0.2702 c.) P(exactly 18 will be cured) = $\frac{20!}{18!2!}$$0.9^{18}$$(1-0.9)^{2}$ = 0.2852 d.) 90% of 20 = 18 P(exactly 90% will be cured) = P(exactly 18 will be cured) P(exactly 90% will be cured) = $\frac{20!}{18!2!}$$0.9^{18}$$(1-0.9)^{2}$ = 0.2852
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