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?
To calculate the probability of receiving 3 or 4 defective units in a shipment of 15 units, we can use the binomial probability formula.
The binomial probability formula is:
P(x) = (nCx) * p^x * q^(n-x)
where:
- P(x) is the probability of getting x successes.
- n is the total number of trials or units in this case (15 units).
- x is the specific number of successes we want (3 or 4 defective units).
- p is the probability of getting a defective unit, which is given as 5% or 0.05 as a decimal.
- q is the probability of not getting a defective unit, which is 1 - p.
Now, let's calculate the probability of receiving 3 defective units first:
P(3) = (15C3) * (0.05)^3 * (1 - 0.05)^(15-3)
= (15! / (3! * (15-3)!)) * 0.05^3 * (0.95)^12
= (15! / (3! * 12!)) * 0.05^3 * 0.95^12
Next, let's calculate the probability of receiving 4 defective units:
P(4) = (15C4) * (0.05)^4 * (1 - 0.05)^(15-4)
= (15! / (4! * (15-4)!)) * 0.05^4 * (0.95)^11
= (15! / (4! * 11!)) * 0.05^4 * 0.95^11
To find the probability of receiving 3 or 4 defective units, we need to sum up the probabilities of these two events since they are mutually exclusive:
P(3 or 4) = P(3) + P(4)
By substituting the calculated values and performing the arithmetic, you will get the final probability.