A company claims that no more than 5% of a particular product has a defect. In a shipment of 15 units, what is the probability of receiving 3 or 4 defective units?

This cannot be answered the way the question is worded. What you need to know is the probability that a given unit is defective. All you have is a manufacturer's claim of "no more than 5%", with no statistical significance. That could imply a wide range of probailities

If the question were worded that the company claims that 5% of a particuar product has a defect, could this be worked?

Yes, if the company claims that exactly 5% of the particular product has a defect, then we can calculate the probability of receiving 3 or 4 defective units.

To find the probability, we can use the binomial probability formula:

P(X = k) = (n C k) * p^k * (1 - p)^(n - k),

where P(X = k) is the probability of obtaining exactly k successes (in this case, defective units), n is the number of trials (total number of units in the shipment), p is the probability of success (probability of a unit being defective), and (n C k) represents the number of ways to choose k defects out of n units.

If the probability of a unit being defective is exactly 5% (or 0.05), we can substitute the values into the formula:

P(X = 3) = (15 C 3) * 0.05^3 * (1 - 0.05)^(15 - 3),

P(X = 4) = (15 C 4) * 0.05^4 * (1 - 0.05)^(15 - 4).

Calculating these values will give you the probability of receiving 3 or 4 defective units in the shipment.