A manufacturing company has a 2% defect rate and seven items are chosen at random. What is the probability that one will have a defect?

Answer 1

Given the scenario and context, we can first identify that this follows binomial experiment criteria; thus, we'll use binomial probability.

Binomial probability formula: P(X = x) = [n! / (n-X)!X!] * p^X * (1-p)^n-X

where X represents the random variable ("success" - the focus of the study), n is sample space/number of trials, and p is the probability of success.

In this scenario, we're told that the the company "has a 2% defect rate". From that point, we can identify that the random variable, or success (X) is those items that are defective, and therefore the probability of success (p) is 2%, or .02. Then we're told that they are looking at a sample of 7 items to test whether any are defective or not. (Note that in binomial, we have both the "success" and then it's complement, which in this case are those items that are non-defective and well-functioning - the probability of this, what we call "probability of failure", is 98%, since 98% of the company's population of items tend to be non-defective and working.)

With the question being "what is the probability that one will have a defect?", we're being asked to find the binomial probability of exactly one out of the 7 in the sample turning out to be defective. Here's how it will set up in the formula --

X = 1 ; n = 7 ; p = .02

P(X = 1) = [7! / (7-1)!1!] * (.02)¹ * (1-.02)^7-1 = 7 * .02 * (.98)⁶ = .124

Reminder: the first part of the formula that involves the factorial (!) - shown in the brackets - is actually the specific formula for Combination (nCr).

Tip: if you happen to have and utilize the TI-84 calculator (or a high-functioning scientific/graphing calculator like that one), you can also use the built-in function to find the binomial probability for this problem. In the TI-84 specifically, that function would be found by clicking 2nd, then VARS to take you to the DISTR menu, where you'll look for the function binompdf. (Use "pdf" for exact values, and "cdf" for an interval set of values.) The full notation to calculate this problem will be binompdf(7, .02, 1).

From the calculation there, the probability that exactly one item (from the sample of 7) is approximately .124, or 12.4%.

Hope this helps! Happy Learning :)