59 % of U. S. adults have very little confidence in newspapers. You randomly select 10 U. S. adults. Find the probability that the number of U. S. adults who have very little confidence in newspapers is (a) exactly five, (b) at least six, and (c) less than four.

(a) 20.9%

(b) 60.8%

(c) 6.3%

Step-by-step explanation:

Use binomial probability.

P = nCr pʳ qⁿ⁻ʳ

(a)

P = ₁₀C₅ (0.59) ⁵ (1-0.59) ¹⁰⁻⁵

P = 0.209

(b)

P = ₁₀C₆ (0.59) ⁶ (1-0.59) ¹⁰⁻⁶

+ ₁₀C₇ (0.59) ⁷ (1-0.59) ¹⁰⁻⁷

+ ₁₀C₈ (0.59) ⁸ (1-0.59) ¹⁰⁻⁸

+ ₁₀C₉ (0.59) ⁹ (1-0.59) ¹⁰⁻⁹

+ ₁₀C₁₀ (0.59) ¹⁰ (1-0.59) ¹⁰⁻¹⁰

P = 0.608

(c)

P = ₁₀C₀ (0.59) ⁰ (1-0.59) ¹⁰⁻⁰

+ ₁₀C₁ (0.59) ¹ (1-0.59) ¹⁰⁻¹

+ ₁₀C₂ (0.59) ² (1-0.59) ¹⁰⁻²

+ ₁₀C₃ (0.59) ³ (1-0.59) ¹⁰⁻³

P = 0.063