Thread: Binomial Question or Counting Techniques?

Hi,

I have a basic probability question that I just cannot understand..

Three of twenty tyres in a store are defective. Four tyres are randomly selected for inspection. What is the probability that at least one defective tyre will be included?

I can think of two solutions but don't understand why they are different / give different answers:

1) There are 20C4 ways of choosing 4 tyres from 20. There are 17C4 ways of choosing 4 non-defective tyres.
Hence P(at least 1 defective tyre) = 1 - P(four non-defective tyres)
= 1 - 17C4 / 20C4
= 0.5087..

2) P(at least 1 defective tyre) = 1 - 4C0 * (3/20)^0*(17/20)^4
=1-(17/20)^4
=0.47799...

My intuition tells me the first method is correct, using counting techniques. But how come the second method (binomial distribution) does not work? I know i have assumed that the rate of defect is 3/20, is this wrong?

Thanks everyone

The second method is wrong, because the probability is not the same as you pick tyres.

At first, the probability is 17/20, then 16/19, then 15/18, and last, 14/17.

You cannot use a binomial distribution since the events are dependent.

That said, the probability becomes the one in the first case.