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
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