Question

Of all the registered automobiles in a certain state, 15% violate the state emissions standard. Twelve automobiles are selected at random to undergo an emissions test.

a)Find the probability that exactly three of them violate the standard

b)Find the probability that fewer than three of them violate the standard.

c)Find the probability that none of them violate the standard.

Answer 1

Solution

Given that ,

p = 0.15

1 - p = 0.85

n = 12

Using binomial probability formula ,

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

a)

P(X = 3) = ((12! /3 ! (12 - 3)!) * 0.15³ * (0.85)^{12 - 3}

=  ((12! / 3! (9)!) * 0.15³ * (0.85)⁹

= 0.1720

Probability = 0.1720

b)

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

=  ((12! /0 ! (12 - 0)!) * 0.15⁰ * (0.85)^{12 - 0} + ((12! /1 ! (12 - 1)!) * 0.15¹* (0.85)^{12 - 1} + ((12! /2 ! (12 - 2)!) * 0.15² * (0.85)^{12 - 2}

=  ((12! / 0! (12)!) * 0.15⁰ * (0.85)¹² + ((12! / 1! (11)!) * 0.15¹ * (0.85)¹¹ + ((12! / 2! (10)!) * 0.15² * (0.85)¹⁰​​​​​​​

= 0.1422 + 0.3012 + 0.2924

= 0.7385

Probability = 0.7358

c)

P(X = 3) = ((12! /0 ! (12 - 0)!) * 0.15⁰ * (0.85)^{12 - 0}

=  ((12! / 0! (12)!) * 0.15⁰ * (0.85)¹²

= 0.1422

Probability = 0.1422