Bayes Theorem
DAILY_CHALLENGE
Grade None
Question:
A factory has a total of three manufacturing units, $M_1, M_2$, and $M_3$, which produce bulbs independent of each other. The units $M_1, M_2$, and $M_3$ produce bulbs in the proportions of $2 : 2 : 1$, respectively. It is known that $20\%$ of the bulbs produced in the factory are defective. It is also known that, of all the bulbs produced by $M_1$, $15\%$ are defective. Suppose that, if a randomly chosen bulb produced in the factory is found to be defective, the probability that it was produced by $M_2$ is $\dfrac{2}{5}$.
If a bulb is chosen randomly from the bulbs produced by $M_3$, then the probability that it is defective is
Step-by-Step Solution
Key Concept: Applying Bayes' Theorem and the Law of Total Probability to find conditional probabilities of partition events.
Given proportions $2:2:1$ for $M_1, M_2, M_3$:
$$P(M_1) = 0.4, \quad P(M_2) = 0.4, \quad P(M_3) = 0.2$$
Let $D$ be the event of choosing a defective bulb:
- Total defective: $P(D) = 0.20$
- Defective from $M_1$: $P(D | M_1) = 0.15$
- Posterior probability for $M_2$: $P(M_2 | D) = 0.40$
By Bayes' theorem:
$$P(M_2 | D) = \dfrac{P(D | M_2) P(M_2)}{P(D)} \implies 0.40 = \dfrac{P(D | M_2) \cdot 0.40}{0.20} \implies P(D | M_2) = 0.20$$
The total probability of defectives is:
$$P(D) = P(D | M_1) P(M_1) + P(D | M_2) P(M_2) + P(D | M_3) P(M_3)$$
0.20 = (0.15 \times 0.40) + (0.20 \times 0.40) + P(D | M_3) \times 0.20$$
$$0.20 = 0.06 + 0.08 + 0.20 \cdot P(D | M_3)$$
$$0.06 = 0.20 \cdot P(D | M_3) \implies P(D | M_3) = 0.30$$
Correct Answer: