Maths help probability

Question

of all the new vehicles of a certain model that are sold 20% require repairs during the first year of service. A particular dealership sells 12 such vehicles. What is the probability that fewer than four of them require repair? 

a. 0.2054,

b. 0.9274,

c. 0.0726,

d. 0.7946

Answer 1

This problem can be approached with binomial expansion.

Lets call r= repair needed (probability 0.2) n= repair not needed (probality 0.8)

The probability of fewer than 4 need repair applies when precisely either 0,1,2 or 3 cars need repair

Now lets take the binomial expansion of (n+r)^12 which is

n^12r^0 + 12(n^11r^1)+66( n^10r^2) + 220(n^9r^3)...........12(n^1r^11) +n^0r^12 I did not write out all the terms as we are only interested in the first 4 terms where the number of cars needing repair is 0,1,2 or 3

Lets calculate the value of the first 4 terms

n^12r^0= .8^12=.0687

n^11r^1= .8^11 x .2^1 =.2061

n^10r^2 = .8^10 x .2^2 = .2834

n^9r^3 = .8^9 x .2^3 = .2362

The sum of these 4 is .7946