Answer to Question #116182 in Statistics and Probability for Otchere-Peprah Kwaku

Question

A package of 6 fuses are tested where the probability an individual fuse is defective is
0.05. (That is, 5% of all fuses manufactured are defective).
(a) What is the probability one fuse will be defective?
(b) What is the probability at least one fuse will be defective?
(c) What is the probability that more than one fuse will be defective, given that at
least one is defective?

Answer 1

a) P(x=1)=C_n,kp^kq^n-k=(6!/(5!*1!))*0.05¹*0.95⁵=0.232

b) P(x≥1)=1-P(x=0)

P(x=0)=(6!/(6!*0!))*0.05⁰*0.95⁶=0.735

P(x≥1)=1-0.735=0.265

c)P(x>1|x≥1)=P(x>1)/P(x≥1)

P(x>1)=1-P(x=0)-P(x=1)=1-0.735-0.232=0.033

P(x>1|x≥1)=0.033/0.265=0.124