In a toy manufacturing company three machines A,B and C are employed to manufactures toys. Machines A, B and C manufacture 20%, 30% and 50% of the total toys respectively. A quality control officer examined the machines and found that A,B and C produce 2%, 3% and 5% defectives of the total output. A toy is selected at random and is found to be defective. What are the probabilities that this toy came from machine A, B and C respectively
0.1
I get:
.02(.2) + .03(.3) + .05(.5) = .038
To find the probabilities that the defective toy came from machines A, B, and C respectively, we can use Bayes' Theorem.
Let's denote the events as follows:
A: The toy is from machine A.
B: The toy is from machine B.
C: The toy is from machine C.
D: The toy is defective.
We are given the following probabilities:
P(A) = 0.20 (machine A's production percentage)
P(B) = 0.30 (machine B's production percentage)
P(C) = 0.50 (machine C's production percentage)
P(D|A) = 0.02 (defective toys produced by machine A)
P(D|B) = 0.03 (defective toys produced by machine B)
P(D|C) = 0.05 (defective toys produced by machine C)
Now, we need to calculate the posterior probabilities P(A|D), P(B|D), and P(C|D).
Using Bayes' Theorem:
P(A|D) = (P(D|A) * P(A)) / P(D)
P(B|D) = (P(D|B) * P(B)) / P(D)
P(C|D) = (P(D|C) * P(C)) / P(D)
To find P(D), we can apply the Law of Total Probability:
P(D) = P(D|A) * P(A) + P(D|B) * P(B) + P(D|C) * P(C)
Calculating the values:
P(D) = (0.02 * 0.20) + (0.03 * 0.30) + (0.05 * 0.50) = 0.01 + 0.009 + 0.025 = 0.044
Now, we can substitute the values into the equations to find the probabilities:
P(A|D) = (0.02 * 0.20) / 0.044 ≈ 0.0909 (approximately 9.09%)
P(B|D) = (0.03 * 0.30) / 0.044 ≈ 0.2045 (approximately 20.45%)
P(C|D) = (0.05 * 0.50) / 0.044 ≈ 0.5682 (approximately 56.82%)
Therefore, the probabilities that the defective toy came from machines A, B, and C respectively are approximately 9.09%, 20.45%, and 56.82%.
To solve this problem, we can use Bayes' theorem, which allows us to calculate the conditional probability of an event given some evidence.
Let's define the following events:
- A: The toy is from machine A
- B: The toy is from machine B
- C: The toy is from machine C
- D: The toy is defective
We can calculate the probabilities of A, B, and C, which are the prior probabilities, based on the percentage of toys produced by each machine:
P(A) = 20%
P(B) = 30%
P(C) = 50%
We also have the probabilities of D given A, B, and C, which are the conditional probabilities. They represent the likelihood of a toy being defective given that it came from a specific machine:
P(D|A) = 2%
P(D|B) = 3%
P(D|C) = 5%
The goal is to calculate the probabilities of A, B, and C given that the toy selected is defective. We can calculate these probabilities using Bayes' theorem:
P(A|D) = P(D|A) * P(A) / P(D)
P(B|D) = P(D|B) * P(B) / P(D)
P(C|D) = P(D|C) * P(C) / P(D)
To find P(D), we can use the law of total probability to consider all possibilities:
P(D) = P(D|A) * P(A) + P(D|B) * P(B) + P(D|C) * P(C)
Now, let's calculate the probabilities:
P(D) = (2% * 20%) + (3% * 30%) + (5% * 50%)
= 0.4% + 0.9% + 2.5%
= 3.8%
P(A|D) = (2% * 20%) / 3.8%
= 0.4% / 3.8%
≈ 0.1053 or 10.53%
P(B|D) = (3% * 30%) / 3.8%
= 0.9% / 3.8%
≈ 0.2368 or 23.68%
P(C|D) = (5% * 50%) / 3.8%
= 2.5% / 3.8%
≈ 0.6579 or 65.79%
Therefore, the probabilities that the defective toy came from machines A, B, and C are approximately 10.53%, 23.68%, and 65.79%, respectively.