A company manufactured a batch of CDs and 3% are defective. If 10 are selected and tested, what is the probability that at least 1 CD is defective?

Answer 1

This is a binomial probability problem. We have a fixed probability of a single success, and each trial is independent of every other trial. The formula is as follows:

P(X = x) = (_nC_x) • p^x • (1 - p)^{n - x}

p = 0.03

n = 10

What we need is the probability that at least 1 is defective. This means the sum of P(X = 1), P(X = 2), ... up to P(X = 10). Calculating all of these and adding them up is a lot of work. Happily, there is a quicker way.

We can easily find the probability that none are defective, which is P(X = 0). A little logic should tell you that this is actually the opposite, or complement, of what we're looking for. So all we have to do is find this probability, and then subtract from 1.

P(X = 0) = (₁₀C₀) • 0.03⁰ • 0.97¹⁰ ≈ 0.737.

The opposite of this is 1 - 0.737 ≈ 0.263.