In a bulb factory, machines P, Q and R manufacture respectively 25%, 35% and 40% of the total. Of their output 5, 4 and 2 percent respectively are defective bulbs. A bulb is drawn at random and it is found to be defective. What is the probability that it was manufactured by machine Q?
0.41
Understanding the Defective Bulb Probability Problem
This problem asks us to find the probability that a randomly selected defective bulb was manufactured by a specific machine, given the production share of each machine and their respective defective rates. This type of problem is a classic application of Bayes' Theorem in probability.
Defining Events and Probabilities
Let's define the events:
- $M_P$: The event that a bulb is manufactured by machine P.
- $M_Q$: The event that a bulb is manufactured by machine Q.
- $M_R$: The event that a bulb is manufactured by machine R.
- $D$: The event that a bulb is defective.
We are given the following probabilities based on the factory's production data:
- Probability of a bulb being from machine P, $P(M_P) = 25\% = 0.25$.
- Probability of a bulb being from machine Q, $P(M_Q) = 35\% = 0.35$.
- Probability of a bulb being from machine R, $P(M_R) = 40\% = 0.40$.
We are also given the conditional probabilities of a bulb being defective, given the machine that manufactured it:
- Probability of a defective bulb from machine P, $P(D | M_P) = 5\% = 0.05$.
- Probability of a defective bulb from machine Q, $P(D | M_Q) = 4\% = 0.04$.
- Probability of a defective bulb from machine R, $P(D | M_R) = 2\% = 0.02$.
We need to find the probability that a bulb was manufactured by machine Q, given that it is defective. This is the conditional probability $P(M_Q | D)$.
Applying Bayes' Theorem for Conditional Probability
Bayes' Theorem is used to find the probability of an event based on prior knowledge of conditions that might be related to the event. The formula for finding $P(M_Q | D)$ is:
$$P(M_Q | D) = \frac{P(D | M_Q) \times P(M_Q)}{P(D)}$$
To use this formula, we first need to calculate the total probability of drawing a defective bulb, $P(D)$. We can do this using the Law of Total Probability, which states that the total probability of an event is the sum of its probabilities under each possible condition.
Calculating the Total Probability of a Defective Bulb
The total probability of a bulb being defective, $P(D)$, is the sum of the probabilities of it being defective and coming from each machine:
$$P(D) = P(D | M_P) \times P(M_P) + P(D | M_Q) \times P(M_Q) + P(D | M_R) \times P(M_R)$$
Let's substitute the given values into this formula:
$$P(D) = (0.05 \times 0.25) + (0.04 \times 0.35) + (0.02 \times 0.40)$$
Calculate each term:
- $0.05 \times 0.25 = 0.0125$ (Probability of being defective AND from machine P)
- $0.04 \times 0.35 = 0.0140$ (Probability of being defective AND from machine Q)
- $0.02 \times 0.40 = 0.0080$ (Probability of being defective AND from machine R)
Now, sum these probabilities to get the total probability of a defective bulb:
$$P(D) = 0.0125 + 0.0140 + 0.0080 = 0.0345$$
So, the total probability of selecting a defective bulb is 0.0345.
Calculating the Probability of Defective Bulb from Machine Q
Now we have all the components to calculate $P(M_Q | D)$ using Bayes' Theorem:
$$P(M_Q | D) = \frac{P(D | M_Q) \times P(M_Q)}{P(D)}$$
Substitute the calculated and given values:
$$P(M_Q | D) = \frac{0.04 \times 0.35}{0.0345}$$
$$P(M_Q | D) = \frac{0.0140}{0.0345}$$
Now, perform the division:
$$P(M_Q | D) \approx 0.405797...$$
Rounding this value to two decimal places, we get approximately 0.41.
Comparing with Options
Let's compare our calculated probability, approximately 0.41, with the given options:
- Option 1: 0.41
- Option 2: 0.55
- Option 3: 0.45
- Option 4: 0.31
Our calculated value matches Option 1.
Revision Table: Key Probability Concepts
| Concept | Description | Formula (Example with D and M) |
|---|---|---|
| Conditional Probability | The probability of an event occurring given that another event has already occurred. | $P(A|B) = \frac{P(A \cap B)}{P(B)}$ |
| Bayes' Theorem | Relates the conditional probability of events. Useful for updating probabilities based on new evidence. | $P(M|D) = \frac{P(D|M) \times P(M)}{P(D)}$ |
| Law of Total Probability | States that the probability of an event is the sum of its probabilities under all mutually exclusive and exhaustive conditions. | $P(D) = \sum_i P(D|M_i) \times P(M_i)$ |
Additional Information: Probability and Manufacturing Defects
Understanding probability is crucial in manufacturing and quality control. By analyzing defective rates and production volumes, companies can identify potential issues with specific machines or processes. This problem illustrates how conditional probability and Bayes' Theorem can be used in reverse to determine the likely cause (which machine) given an observed outcome (a defective bulb).
Probability distributions are often used to model the number of defects in a batch or the time between defects. The binomial distribution, for example, can model the number of defective items in a fixed sample size, assuming a constant probability of defect.
In real-world scenarios, these calculations help optimize production processes, improve quality, and manage resources effectively. Monitoring these probabilities over time can also indicate whether machine performance is changing.
Important Questions from Conditional Probability
A and B are two events such that A̅ and B̅ are mutually exclusive. If P(A) = 0.5 and P(B) = 0.6, then what is the value of P(A|B)?
For two dependent events A and B, it is given that P(A) = 0.2 and P(B) = 0.5. If A ⊆ B, then the values of conditional probabilities P(A|B) and P(B|A) are respectively
Let X1 and X2 be independent random variables each having geometric distribution qk p ; k = 0, 1, 2, …. Then the conditional distribution of X1 given X1 + X2 is
In a batch Of 100 resistors Of 1kiloohm resistance, 80 numbers are within the required tolerance values and II numbers are below the required tolerance values, the remaining are above the required tolerance values. If two resistors are drawn one after the other without replacement, the probability of the first one drawn is below and the second one drawn is above the required tolerance value is
A computer producing factory has only two plants T1 and T2. Plant T1 produces 20% and plant T2 produces 80% of the total computers produced. 7% of the computers produced in the factory turn out to be defective. It is known that P(computer turns out to be defective given that it is produced in plant T2) = 10 × P(computer turns out to be defective given that it is produced in plant T1). A computer produced in the factory is randomly selected and it does not turn out to be defective, then the probability that it is produced in plant T2 is:
