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?
You can probably use a binomial probability function for this problem.
P(x) = (nCx)(p^x)[q^(n-x)]
n = 15
x = 3, 4
p = .05
q = 1 - p = .95
Find P(3), P(4).
P(3) = (15C3)(.05^3)(.95^12)
P(4) = (15C4)(.05^4)(.95^11)
Finish the calculations; then add together for your probability.
To calculate the probability of receiving 3 or 4 defective units out of a shipment of 15 units, we can use the binomial probability function. The formula is:
P(x) = (nCx)(p^x)(q^(n-x))
Where:
n = total number of units in the shipment (15 in this case)
x = number of defective units (3 or 4)
p = probability of a unit being defective (0.05 or 5%)
q = probability of a unit not being defective (1 - p)
Let's calculate P(3) first:
P(3) = (15C3)(0.05^3)(0.95^12)
To determine (15C3), we calculate the combination value using the formula:
(15C3) = 15! / (3!(15-3)!)
= (15 * 14 * 13) / (3 * 2 * 1)
= 455
Now, substitute the values into the binomial probability formula:
P(3) = (455)(0.05^3)(0.95^12)
= 0.0002145
Next, let's calculate P(4):
P(4) = (15C4)(0.05^4)(0.95^11)
Using the same method as above, we find:
(15C4) = 15! / (4!(15-4)!)
= (15 * 14 * 13 * 12) / (4 * 3 * 2 * 1)
= 1365
Substituting values:
P(4) = (1365)(0.05^4)(0.95^11)
= 0.001057
Finally, to find the probability of receiving either 3 or 4 defective units, we add the two probabilities together:
P(3 or 4) = P(3) + P(4)
= 0.0002145 + 0.001057
= 0.0012715
Therefore, the probability of receiving 3 or 4 defective units in a shipment of 15 units is approximately 0.0012715, or 0.13%.