Question

Forty-four percent of US adults have little confidence in their cars. You randomly select twelve US adults. Find the probability that the number of US adults who have little confidence in their cars is (1) exactly six and then find the probability that it is (2) more than 7.

Answer 1

Assuming the probability of success 'p' i.e the probability of US adults having little confidence in their cars = 45%

p = 0.45

Hence, the probability of failure q = 1- p = 1 - 0.45

q =0.55

Hence we can find the probability P(x) using the binomial distribution formula,

$P(x) = \frac {n!}{(n-x)!*x!}*p^x*q^{n-x}$

where n = total number of random selection of people

and x = total number of success required

CASE 1)

n = 12, x = 6

Hence P(x) = $\frac {12!}{(12 - 6)!*6!}*0.45^6*0.55^6=0.21238$

Hence, the probability that the number of US adults who have little confidence in their cars is exactly 6 from a random selection of 12 people = 0.21238

CASE 2)

n = 12, x > 7 (means sum of all probabilities for x = 8, 9, 10, 11 & 12)

Hence,

P(x=8) = $\frac {12!}{(12 - 8)!*8!}*0.45^8*0.55^4=0.07616$

P(x=9) = $\frac {12!}{(12 - 9)!*9!}*0.45^9*0.55^3=0.02769$

P(x=10) = $\frac {12!}{(12 - 10)!*10!}*0.45^{10}*0.55^2=0.006798$

P(x=11) = $\frac {12!}{(12 - 11)!*11!}*0.45^{11}*0.55^1=0.0010113$

P(x=12) = $\frac {12!}{(12 - 12)!*12!}*0.45^{12}*0.55^0=0.0000689525$

Thus,

P(x>7) = P(x=8) + P(x=9) + P(x=10) + P(x=11) + P(x=12)

P(x>7) = 0.7616 + 0.02769 + 0.006799 + 0.0010113 + 0.0000689525 = 0.111728

Hence, the probability that the number of US adults who have little confidence in their cars is more than 7 from a random selection of 12 people = 0.111728