Question

A box contains 17 ​transistors, 3 of which are defective. If 3 are selected at​ random, find the probability that

a. All are defective.
b. None are defective.

Answer 1

b none are defective

Answer 2

Use Binomial Theorem

`P(x=k) = (nCk) p^k (1-p)^(n-k)`
Where k is number that are defective, n is total selected (3), and p is probability that a transistor is defective (3/17).

a) All are defective means k = 3

`P = (3C3) ((3)/(17))^3 ((14)/(17))^0 = ((3)/(17))^3 = 0.0055`

b) None are defective means k = 0

`P = (3C0) ((3)/(17))^0 ((14)/(17))^3 = ((14)/(17))^3 = 0.5585`