Question

66​% 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

X ~ Binomial (n,p)

Where n = 10 , p = 0.66

Binomial probability distribution is

P (X) = ⁿC_x p^x (1 - p)^n-x

a)

P (X = 5) = ¹⁰C₅ 0.66⁵ 0.34⁵

= 0.1434

b)

P( X >= 6) = P (X = 6) + P (X = 7) + P (X = 8) + P (X = 9) + P (X = 10)

=  ¹⁰C₆ 0.66⁶ 0.34⁴ +¹⁰C₇ 0.66⁷ 0.34³ +¹⁰C₈ 0.66⁸ 0.34² +¹⁰C₉ 0.66⁹ 0.34 +¹⁰C₁₀ 0.66¹⁰ 0.34⁰

= 0.7730

c)

P( X < 4) = P (X <= 3)

= P (X = 0) +P (X = 1) +P (X = 2) +P (X = 3)

= ¹⁰C₀ 0.66⁰ 0.34¹⁰ +¹⁰C₁ 0.66¹ 0.34⁹ +¹⁰C₂ 0.66² 0.34⁸ +¹⁰C₃ 0.66³ 0.34⁷

= 0.0220