An automobile tire dealer buys tires in lots of 750. His experience is that 1% of all new tires purchased must be replaced within 30 days as they are defective. Find the probability that in a lot of 750 tires, there will be a) No defective tires ( 3 Marks) b) Less than two defective tires

a. 5.33 × 10⁻⁴ b. 4.568 × 10⁻³

Step-by-step explanation:

a. This situation represents a binomial probability. Let p = probability that tire is defective = 1% = 0.01, then q = probability that tire is not defective = 1 - p = 1 -   0.01 = 0.99 = 99%.

Since there are 750 tires, the probablility that no tire is defective is P = ⁷⁵⁰C₀q⁷⁵⁰p⁰

= (1) (0.99)⁷⁵⁰(0.01)⁰

= 5.33 × 10⁻⁴

b. The probability that less than two tires will be defective P'

So, P' = ⁷⁵⁰C₀q⁷⁵⁰p⁰ + ⁷⁵⁰C₁q⁷⁴⁹p¹

= (1) (0.99)⁷⁵⁰(0.01)⁰ + (750)(0.99)⁷⁴⁹(0.01)¹

= 5.33 × 10⁻⁴ + 0.004035

= 45.68 × 10⁻⁴

=  4.568 × 10⁻³

i think is g(1)=-1.

hope that i can