A manufacturer of cotter pin know that 5% of his product is defective. If he sells cotter pins in boxes of 100 pins find the probability that
a) 10 pins will be defective in the box
b) Not more than 10 pins are defective in a box
prob(defect) = .05
prob(no defect) = .95
a) I read that as exactly 10 out of the 100
= C(100,10) (.05)^10 (.95)^90
= appr .0167, according to my calculator
b) unreasonable question, you would have to do
prob(exactly 0 defective) + prob(1 effective) + ...+ prob(10 defective)
If you feel like it, repeat the pattern I used in a)
To find the probabilities, we can use the binomial probability formula:
P(x) = C(n, x) * p^x * q^(n-x)
where:
P(x) is the probability of x successes,
C(n, x) is the number of combinations of n items taken x at a time,
p is the probability of success in a single trial, and
q is the probability of failure in a single trial (1-p).
In this case:
n = 100 (number of cotter pins in the box)
p = 0.05 (probability of a defective pin)
q = 1 - p = 1 - 0.05 = 0.95
a) Probability that 10 pins will be defective in the box:
P(10) = C(100, 10) * (0.05)^10 * (0.95)^(100-10)
Calculating this:
P(10) = (100! / (10! * (100-10)!)) * (0.05)^10 * (0.95)^90
Using a calculator or software to compute the combination (C) and exponentiation, we find:
P(10) ≈ 0.12918
Therefore, the probability that exactly 10 pins will be defective in the box is approximately 0.12918.
b) Probability that not more than 10 pins are defective in a box:
To find this probability, we need to calculate the probabilities for 0, 1, 2, 3, ..., and 10 defective pins and sum them up.
P(not more than 10) = P(0) + P(1) + P(2) + ... + P(10)
Using the same formula as before but summing up the individual probabilities, we get:
P(not more than 10) = Σ P(x) (where x = 0 to 10)
Calculating this:
P(not more than 10) = P(0) + P(1) + P(2) + ... + P(10)
Using a calculator or software, we can calculate each term and sum them:
P(not more than 10) ≈ 0.999999994 (approximately 1)
Therefore, the probability that not more than 10 pins are defective in a box is approximately 1.
To find the probability in both cases, we will use the binomial probability formula:
P(x) = C(n, x) * p^x * (1-p)^(n-x)
Where:
P(x) is the probability of getting x successful outcomes (defective pins in this case),
n is the total number of trials (number of cotter pins in the box),
p is the probability of success (probability of getting a defective pin),
C(n, x) is the combination formula (nCx) which calculates the number of ways x successes can occur in n trials.
a) Probability of 10 pins being defective:
Given that 5% of the cotter pins are defective, the probability of getting a defective pin is p = 0.05.
n = 100 (number of cotter pins in the box)
x = 10 (number of defective pins we want to find the probability for)
Using the binomial probability formula, we have:
P(10) = C(100, 10) * 0.05^10 * (1-0.05)^(100-10)
To calculate this, we need to evaluate the combination formula and perform the necessary calculations.
b) Probability of not more than 10 pins being defective:
This can be calculated by finding the sum of probabilities for 0, 1, 2, 3, ..., 10 defective pins.
P(not more than 10) = P(0) + P(1) + P(2) + ... + P(10)
We can use the binomial probability formula as explained above for each value of x (defective pins), and then sum up the probabilities.
I hope this helps you understand how to calculate the probabilities for these scenarios involving defective cotter pins. Let me know if you have any further questions!