Question

A manufacturer produces a large number of toasters. From past experience, the manufacturer knows that approximately 1% are defective. In a quality control procedure, we randomly select 50 toasters for testing. We want to determine the probability that no more than one of these toasters is defective.

Answer 1

Answer:

The probability is 0.9106

Step-by-step explanation:

The variable that says the number of defective toasters follows a binomial distribution, where we have n identical and independent events (50 toasters) with a probability p of success (1% are defective) and a probability 1-p of fail (99% are not defective). So the probability that x toasters from the 50 are defective is:

$P(x) = \frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x} \\P(x) = \frac{50!}{x!(50-x)!}*0.01^{x}*(1-0.01)^{50-x}$

Then, the probability P that no more than one of these toasters is defective is:

P = P(0) + P(1)

So, P(0) and P(1) are calculated as:

$P(0)=\frac{50!}{0!(50-0)!}*0.01^{0}*(1-0.01)^{50-0}=0.6050$

$P(1)=\frac{50!}{1!(50-1)!}*0.01^{1}*(1-0.01)^{50-1}=0.3056$

Finally, P is equal to:

P = 0.6050 + 0.3056 = 0.9106