Question

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

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
$P = (3C3) (\frac{3}{17})^3 (\frac{14}{17})^0 = (\frac{3}{17})^3 = 0.0055$

b) None are defective means k = 0
$P = (3C0) (\frac{3}{17})^0 (\frac{14}{17})^3 = (\frac{14}{17})^3 = 0.5585$