Question: Let $X$ represent the event that a component is from Factory $X$, $Y$ represent the event that a component is from Factory $Y$, and $Z$ represent the event that a component is from Factory $Z$. Let $D$ represent the event that a component is defective.
a) Find the probability that a randomly selected component is non-defective and comes from Factory $X$.
b) Find the probability that a randomly selected component is defective.
c) Given that a randomly selected component is defective, find the probability that it came from Factory $Y$.
1. **Problem Statement:**
We have three factories $X$, $Y$, and $Z$ producing components. $D$ is the event that a component is defective.
We want to find:
a) $P(\text{non-defective} \cap X)$
b) $P(D)$
c) $P(Y|D)$
2. **Formulas and Rules:**
- Probability of non-defective from $X$ is $P(X \cap D^c) = P(X) - P(X \cap D)$.
- Total defective probability is $P(D) = P(X \cap D) + P(Y \cap D) + P(Z \cap D)$.
- Conditional probability formula: $P(Y|D) = \frac{P(Y \cap D)}{P(D)}$.
3. **Assumptions:**
Since probabilities or defect rates are not given, let's assume the following example data for illustration:
- $P(X) = 0.3$, $P(Y) = 0.5$, $P(Z) = 0.2$
- Defective rates: $P(D|X) = 0.02$, $P(D|Y) = 0.03$, $P(D|Z) = 0.05$
4. **Calculations:**
**a) Probability non-defective and from $X$:**
$$
P(X \cap D^c) = P(X) \times (1 - P(D|X)) = 0.3 \times (1 - 0.02) = 0.3 \times 0.98 = 0.294
$$
**b) Probability defective:**
$$
P(D) = P(X)P(D|X) + P(Y)P(D|Y) + P(Z)P(D|Z) = 0.3 \times 0.02 + 0.5 \times 0.03 + 0.2 \times 0.05
$$
$$
= 0.006 + 0.015 + 0.01 = 0.031
$$
**c) Probability component came from $Y$ given defective:**
$$
P(Y|D) = \frac{P(Y \cap D)}{P(D)} = \frac{P(Y)P(D|Y)}{P(D)} = \frac{0.5 \times 0.03}{0.031} = \frac{0.015}{0.031} \approx 0.4839
$$
5. **Summary:**
- $P(\text{non-defective} \cap X) = 0.294$
- $P(D) = 0.031$
- $P(Y|D) \approx 0.4839$
These results show how to combine total probability and conditional probability to solve the problem.