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:(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)¹⁰⁻⁵P = 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)¹⁰⁻¹⁰P = 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)¹⁰⁻³P = 0.063