How to find the probability

Your manufacturing plant produces air bags, and it is known that 30% of them are defective. Five air bags are tested.

(a) Find the probability that 5 of them are defective.

(b) Find the probability that at least 2 of them are defective.

The number N of defective air bags tested out of five is a discrete random variable, taking on the values n = {0, 1, 2, ... , 5}.

The probability P(N = n) follows a binomial distribution so that:

P(N = n) = 5Cn p^n (1 - p)^(5 - n)

... where p = 30% = 0.3 = the probability that any particular bag will be found to be defective, so:

P(N = n) = (5Cn) (0.3)^n (0.7)^(5 - n)


PART A

The probability P(N = 5) that all five are defective is:

P(N = 5) = (5C5) (0.3)^5 (0.7)^0
= (0.3)^5
= 0.00243


PART B

The probability P(N ≥ 2) that at least 2 are defective is the same as the complement of the probability that exactly 0 or exactly 1 are defective. That is,

P(N ≥ 2) = 1 - (P(N = 0) + P(N = 1))
= 1 - ((5C0) (0.3)^0 (0.7)^5 + (5C1) (0.3)^1 (0.7)^4)
= 1 - (0.16807 + 0.36015)
= 0.47178