Question

2. A company is ordering 15 tires from a tire factory. What is the probability that the order has at least one defective tire, given that a) The tire factory has in stock 100 tires and five of them are defective. b) The probability of producing a defective tire is 5% and tires are produced independently.

Answer 1

2)a) P(at least one is defective) = 1 - P(no one is defective)

                                                  = 1 - (95C15/100C15)

                                                   $\dpi{120} = 1 - (\frac{\frac{95!}{15! * 80!}}{\frac{100!}{15! * 85!}})$

                                                   $\dpi{100} = 1 - (\frac{95 ! * 85!}{80! * 100!})$

                                                   = 1 - 0.4357 = 0.5643

b) p = 0.05

    n = 15

It is a binomial distribution.

P(X = x) = nCx * p^x * (1 - p)^{n - x}

P(X > 1) = 1 - P(X < 1)

               = 1 - P(X = 0)

               = 1 - (15C0 * (0.05)^0 * (0.95)^15)

               = 1 - 0.4633

               = 0.5367