A stationary store has decided to accept a large shipment of ball-point pens if an inspection of 20 randomly selected pens yields no more than two defective pens.
a) find the probability that this shipment is accepted if 5% of the total shipment is defective.
b) find the probability that this shipment is not accepted if 15% of the total shipment is defective.
a) The probability that this shipment is accepted if 5% of the total shipment is defective is 0.841.
b) The probability that this shipment is not accepted if 15% of the total shipment is defective is 0.093.
a) To find the probability that the shipment is accepted, we need to calculate the probability of having no more than two defective pens in a sample of 20, given that 5% of the total shipment is defective.
Let's use the binomial probability formula:
P(x) = C(n, x) * p^x * q^(n-x)
Where P(x) is the probability of having x defective pens, n is the sample size (20), p is the probability of a pen being defective (0.05), q is the probability of a pen not being defective (1 - p = 0.95), and C(n, x) is the combination formula for choosing x items from a set of n items.
For no more than two defective pens, we need to calculate the probability of having 0, 1, or 2 defective pens, and sum them up:
P(0 defective) = C(20, 0) * (0.05^0) * (0.95^20)
P(1 defective) = C(20, 1) * (0.05^1) * (0.95^19)
P(2 defective) = C(20, 2) * (0.05^2) * (0.95^18)
Now, we can calculate the total probability:
P(shipment accepted) = P(0 defective) + P(1 defective) + P(2 defective)
Let's plug these values into a calculator to find the final probability.
b) Similarly, to find the probability that the shipment is not accepted, we need to calculate the probability of having more than two defective pens in a sample of 20, given that 15% of the total shipment is defective.
P(shipment not accepted) = 1 - P(shipment accepted)
Following the same steps as in part a, we can calculate the probability of having 3, 4, ..., 20 defective pens and subtract this from 1 to find the final probability.
I hope this helps! Let me know if you have any further questions.
To find the probability in both cases, we will use the concept of the binomial distribution. The binomial distribution is used when there are two possible outcomes (defective or not defective) and each trial (inspection of a pen) is independent of each other, with the same probability of success (defectiveness) for each trial.
a) Probability of accepting the shipment if 5% of the total shipment is defective:
Let's denote the probability of a pen being defective as p = 0.05.
And the probability of a pen not being defective as q = 1 - p = 0.95.
Now, we need to find the probability of selecting no more than two defective pens in a sample of 20 pens.
We can calculate this using the binomial distribution formula:
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
Where X represents the number of defective pens in the sample.
P(X = x) = (nCk) * p^k * q^(n-k)
Where n is the total number of trials (number of pens inspected), k is the number of successful trials (number of defective pens), p is the probability of success (defectiveness), and q is the probability of failure (non-defectiveness).
Substituting the values into the formula:
P(X = 0) = (20C0) * (0.05^0) * (0.95^20)
P(X = 1) = (20C1) * (0.05^1) * (0.95^19)
P(X = 2) = (20C2) * (0.05^2) * (0.95^18)
Calculating these probabilities and summing them up will give us the probability of accepting the shipment if 5% of the total shipment is defective.
b) Probability of not accepting the shipment if 15% of the total shipment is defective:
Using the same approach, we can find the probability of selecting more than two defective pens in a sample of 20 pens.
Since we need to calculate P(X > 2), we can find this by subtracting P(X ≤ 2) from 1.
P(X > 2) = 1 - P(X ≤ 2)
By calculating P(X ≤ 2) as described above and subtracting it from 1, we can find the probability of not accepting the shipment if 15% of the total shipment is defective.