simple probability

awilhelm wrote:A car dealer is testing six of his cars to determine if they run properly. The probability that they are defective follows a binomial distribution. The probability of a car being defective is .1. What is the probability that exactly two cars will be defective?

Thanks!

6C2 (.1)^2 (.9)^4

For Binomial Probability Use

NCr x p^r (1-p)^( n-r) where N is number of items and r is number of successes, in this case number of defective cars. NC2 is 2 out of 6 cars, the normal combination thing.

p is probability of success= defective
1-p is probability of failure= not defective

Hope this helps.

We want two cars to be defective and the rest 4 to be perfect

So the probability is 0.1 x 0.1 x 0.9 x 0.9 x 0.9 x 0.9

But two cars can be any 2 from the available 6.

So the prob is 6C2 x 0.1^2 x 0.9^4