
A computer manufacturer uses chips from three sources. Chips from sources A, B and C
are defective with probabilities 0.005, 0.001 and 0.01, respectively. The proportions of chips
from A, B and C are 0.5, 0.1 and 0.4 respectively. If a randomly selected chip is found to
be defective, find
(a) the probability that the chips are from A.
(b) the probability that the chips are from B.
(c) the probability that the chips are from 

We are given the following information about defective chips and their sources:

- The probability that a chip from:
  - Source A is defective: \( P(D|A) = 0.005 \)
  - Source B is defective: \( P(D|B) = 0.001 \)
  - Source C is defective: \( P(D|C) = 0.01 \)
  
- The proportions of chips from each source:
  - Proportion from A: \( P(A) = 0.5 \)
  - Proportion from B: \( P(B) = 0.1 \)
  - Proportion from C: \( P(C) = 0.4 \)

We are asked to find the probability that a defective chip comes from each source. These are conditional probabilities, and we will use **Bayes' theorem** to calculate them.

### Bayes' Theorem:
Bayes' theorem states:
\[
P(A|D) = \frac{P(D|A) P(A)}{P(D)}
\]
where:
- \( P(A|D) \) is the probability that the chip is from source A given that it is defective.
- \( P(D|A) \) is the probability of a defective chip from source A.
- \( P(A) \) is the proportion of chips from source A.
- \( P(D) \) is the total probability that a chip is defective.

Similarly, we can calculate \( P(B|D) \) and \( P(C|D) \).

### Step 1: Calculate \( P(D) \) (Total Probability of Defective Chip)
Using the law of total probability:
\[
P(D) = P(D|A) P(A) + P(D|B) P(B) + P(D|C) P(C)
\]
Substituting the given values:
\[
P(D) = (0.005 \times 0.5) + (0.001 \times 0.1) + (0.01 \times 0.4)
\]
\[
P(D) = 0.0025 + 0.0001 + 0.004 = 0.0066
\]

### Step 2: Calculate \( P(A|D) \) (Probability that defective chip is from A)
Now using Bayes' theorem for source A:
\[
P(A|D) = \frac{P(D|A) P(A)}{P(D)} = \frac{0.005 \times 0.5}{0.0066}
\]
\[
P(A|D) = \frac{0.0025}{0.0066} \approx 0.3788
\]

### Step 3: Calculate \( P(B|D) \) (Probability that defective chip is from B)
\[
P(B|D) = \frac{P(D|B) P(B)}{P(D)} = \frac{0.001 \times 0.1}{0.0066}
\]
\[
P(B|D) = \frac{0.0001}{0.0066} \approx 0.0152
\]

### Step 4: Calculate \( P(C|D) \) (Probability that defective chip is from C)
\[
P(C|D) = \frac{P(D|C) P(C)}{P(D)} = \frac{0.01 \times 0.4}{0.0066}
\]
\[
P(C|D) = \frac{0.004}{0.0066} \approx 0.6061
\]

### Final Results:
- (a) The probability that the defective chip is from source A: \( P(A|D) \approx 0.3788 \).
- (b) The probability that the defective chip is from source B: \( P(B|D) \approx 0.0152 \).
- (c) The probability that the defective chip is from source C: \( P(C|D) \approx 0.6061 \).

Would you like more details or have any questions?

Here are 5 related questions to consider:
1. How would the probabilities change if the proportion of chips from source B increased?
2. Can Bayes' theorem be applied if we had additional sources of chips?
3. What happens to the total defective probability if one source improves its manufacturing process?
4. How would you compute the probability if we knew the exact number of chips produced?
5. What assumptions does Bayes' theorem rely on in this scenario?

**Tip**: Bayes' theorem is extremely useful when updating probabilities based on new information, such as determining the likely cause of an outcome given multiple potential sources.

A computer manufacturer uses chips from three sources. Chips from sources A, B, and C are defective with probabilities 0.005, 0.001, and 0.01, respectively. The proportions of chips from A, B, and C are 0.5, 0.1, and 0.4 respectively. If a randomly selected chip is found to be defective, find (a) the probability that the chips are from A. (b) the probability that the chips are from B. (c) the probability that the chips are from C.

Explore how to calculate the conditional probability that a defective chip comes from one of three sources (A, B, or C) using Bayes' theorem. This detailed solution covers the probability of defective chips and uses the law of total probability. Ideal for students in Grades 10-12.

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