Question

43​% of men consider themselves professional baseball fans. You randomly select 10 men and ask each if he considers himself a professional baseball fan. Find the probability that the number who consider themselves baseball fans is​ (a) exactly​ five, (b) at least​ six, and​ (c) less than four.

Answer 1

Solution

Given that ,

p = 43% = 0.43

1 - p = 1 - 0.43 = 0.57

n = 10

Using binomial probability formula ,

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

a)

P(X = 5) = ((10! / 5! (10 - 5)!) * 0.43⁵ * (0.57)10-5

=  ((10! / 5! (5)!) * 0.43⁵ * (0.57)⁵

= 0.2229

Probability = 0.2229

b)

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

= ((10! / 6! (10-6)!) * 0.43⁶ *   (0.57)^10-6 + ((10! / 7! (10-7)!) * 0.43⁷ *   (0.57)^10-7 + ((10! / 8! (10-8)!) * 0.43⁸ *   (0.57)^10-8 + ((10! / 9! (10-9)!) * 0.43⁹ *   (0.57)^10-9 + ((10! / 10! (10-10)!) * 0.43¹⁰ *   (0.57)^10-10

=  ((10! / 6! (4)!) * 0.43⁶ * (0.57)⁴ + ((10! / 7! (3)!) * 0.43⁷ * (0.57)³ + ((10! / 8! (2)!) * 0.43⁸ * (0.57)² + ((10! / 9! (1)!) * 0.43⁹ * (0.57)1+ ((10! / 10! (0)!) * 0.43¹⁰ * (0.57)⁰

=  0.1401+0.0604+0.0171+0.0029+0.0002

Probability = 0.2207

c)

P(X < 4) = P(X=0)+P(X=1)+P(X=2)+P(X=3)

= (10! / 0! (10-0)!) * 0.43⁰ * (0.57)^10-0 + (10! / 1! (10-1)!) * 0.43¹ * (0.57)^10-1 + (10! /2! (10-2)!) * 0.43² * (0.57)^{10 - 2} + (10! /3 ! (10-3)!) * 0.43⁷ * (0.57)^10-3

⁼ ((10! / 0! (10)!) * 0.43⁰ * (0.57)¹⁰  + ((10! / 1! (9)!) * 0.43¹ * (0.57)⁹  + ((10! / 2! (8)!) * 0.43² * (0.57)⁸  + ((10! /3 ! (7)!) * 0.43³ * (0.57)⁷  

= 0.0036+0.0273+0.0927+0.1865

= 0.3102

Probability = 0.3102