Of all the new vehicles of a certain model that are sold 15

Of all the new vehicles of a certain model that are sold, 15% require repairs to be done under warranty during the first year of service. A particular dealership sells 20 such vehicles. What is the probability that fewer than 4 of them require warranty repairs?

Solution

Of all the new vehicles of a certain model that are sold, 15% require repairs to be done under warranty during the first year of service. A particular dealership sells 20 such vehicles. What is the probability that fewer than 4 of them require warranty repairs?

This is binomial. Number of trials, n = 20. P = .15.

The formula for the probability for the binomial distribution:

P(X=x) = ncx*p^x*(1-p)^(n-x)

Fewer than 4 would be 0, 1, 2, or 3.

We want to find the probability that x is 0, 1,2, or 3.

P(X=0) = 20c0*.15^0*(1-.15)^(20-0) = .038760

P(X=1) = 20c1*.15^1*(1-.15)^(20-1) = .136798

P(X=2) = 20c2*.15^2*(1-.15)^(20-2) = .229338

P(X=3) = 20c3*.15^3*(1-.15)^(20-3) = 0.242829

Add up these four probabilities. Answer is 0.6477.

ANSWER: 0.6477

(Alternatively, you can use a calculator with a binomcdf function. Binomcdf(20, .15, 3) = 0.6477