Question

​ 62% 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

Solution

Given that ,

p = 62% = 0.62

1 - p = 1 - 0.62 = 0.38

n = 10

Using binomial probability formula ,

P(X = x) = ((n! / x! (n - x)!) * p^x * (1 - p)^{n - x}

(a)

x = 5

P(X = 5) = ((10! / 5! (5)!) * 0.62⁵ * (0.38)⁵

= 0.1829

Probability = 0.1829

(b)

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

= ((10! / 6! (4)!) * 0.62⁶ * (0.38)⁴ +  ((10! / 7! (3)!) * 0.62⁷ * (0.38)³ +  ((10! / 8! (2)!) * 0.62⁸ * (0.38)²

+  ((10! / 9! (1)!) * 0.62⁹ * (0.38)¹ +  ((10! / 10! (0)!) * 0.62¹⁰ * (0.38)⁰

= 0.6823

Probability = 0.6823

(c)

P(X < 4) = 1 - P(X 4)

= 1 - {P(X = 4) + P9X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)}

= 1 - { ((10! / 4! (6)!) * 0.62⁴ * (0.38)⁶ + ((10! / 5! (5)!) * 0.62⁵ * (0.38)⁵ + ((10! / 6! (4)!) * 0.62⁶ * (0.38)⁴

+  ((10! / 7! (3)!) * 0.62⁷ * (0.38)³ +  ((10! / 8! (2)!) * 0.62⁸ * (0.38)²

+  ((10! / 9! (1)!) * 0.62⁹ * (0.38)¹ +  ((10! / 10! (0)!) * 0.62¹⁰ * (0.38)⁰ }

= 1 - 0.9587

= 0.0413

Probability = 0.0413