This problem involves finding the probability of a randomly selected integrated circuit (IC) being defective, considering the total ICs purchased from different suppliers and their respective defect rates.

Let's solve step by step.

Step 1: Definitions

Let:

• $A$, $B$, $C$ denote suppliers A, B, and C.
• $D$ denote the event that an IC is defective.

We are given:

• Probability that an IC comes from supplier $A$:
  $P(A) = \frac{1000}{1000 + 2000 + 3000} = \frac{1000}{6000} = \frac{1}{6}$,
• Probability that an IC comes from supplier $B$:
  $P(B) = \frac{2000}{6000} = \frac{1}{3}$,
• Probability that an IC comes from supplier $C$:
  $P(C) = \frac{3000}{6000} = \frac{1}{2}$.

Defect probabilities for each supplier:

• $P(D|A) = 0.05$,
• $P(D|B) = 0.10$,
• $P(D|C) = 0.10$.

Step 2: Use the Law of Total Probability

The total probability of a defect $P(D)$ is given by: $P(D) = P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C).$

Step 3: Substitute Values

Substituting the given values: $P(D) = (0.05)(\frac{1}{6}) + (0.10)(\frac{1}{3}) + (0.10)(\frac{1}{2}).$

Step 4: Simplify

Evaluate each term:

• $(0.05)(\frac{1}{6}) = \frac{0.05}{6} = 0.00833$,
• $(0.10)(\frac{1}{3}) = \frac{0.10}{3} = 0.03333$,
• $(0.10)(\frac{1}{2}) = \frac{0.10}{2} = 0.05$.

Add these probabilities: $P(D) = 0.00833 + 0.03333 + 0.05 = 0.09166.$

Final Answer:

The probability that a randomly selected IC is defective is approximately: $P(D) \approx 0.0917 \, \text{(or 9.17\%)}.$

Would you like a detailed explanation of any step?
Here are five related questions to explore further:

1. How would the probability change if the defect rate for supplier $C$ were different?
2. What is the probability that a defective IC came from supplier $A$?
3. Can we generalize this problem to more suppliers with similar data?
4. How can we apply Bayes' theorem to this problem?
5. What is the role of the law of total probability in solving such problems?

Tip: Always verify your probabilities add up to 1 when partitioning events across disjoint cases.