Question

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.

Answer 1

(a) 20.9%

(b) 60.8%

(c) 6.3%

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