25 buses, cracks on 10 of the buses. 8 buses chosen at random and without replacement. probability that exactly 4 will have visible cracks?

Question

Shortly after being put into service, some buses manufactured by a certain company have developed cracks on the underside of the main frame. Suppose that bus company has 25 of these buses, and that cracks have actually appeared on 10 of the buses. If 8 of these buses are chosen at random and without replacement, what is the probability that exactly 4 of them will have visible cracks?

(A) 0.0038 (B) 0.2650 (C) 0.1538 (D) 0.0166 (E) 0.3218

Answer 1

We are starting out with 25 buses: 10 defective and 15 non-defective.

Let's first figure out the probability of first picking 4 defectives and then 4 non-defectives:

p = (10/25)(9/24)(8/23)(7/22) * (15/21)(14/20)(13/19)(12/18) =

Now let'e count how many different ways the 4 defective buses can be arranged among 8 places:

C(8,4)

P = C(8,4)*p = [ 8! / (4! 4!) ] * 0.003786145204909507 = 70 * 0.003786145204909507

P = 0.2650 = 26.50%