A bike manufacturing factory has two plants P and Q. Plant P manufactures 60 percent of bikes and plant Q manufacture 40 percent. 80 percent of the bikes at plant P and 90 percent of the bikes at plant Q are rated of standard quality. A bike is chosen at random and is found to be of standard quality. What is the probability that it has come from plant P?
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Understanding the Probability Problem: Bayes' Theorem Application
This problem asks for the probability that a bike came from a specific plant (Plant P), given that the bike is of standard quality. This is a classic example of a conditional probability problem that can be solved using Bayes' Theorem.
Defining Events and Given Probabilities
Let's define the events involved in this scenario:
- P: The event that a randomly chosen bike is manufactured at Plant P.
- Q: The event that a randomly chosen bike is manufactured at Plant Q.
- S: The event that a randomly chosen bike is of standard quality.
We are given the following probabilities:
- Probability that a bike is from Plant P, $P(P) = 60\% = 0.60$.
- Probability that a bike is from Plant Q, $P(Q) = 40\% = 0.40$.
- Probability that a bike from Plant P is of standard quality, $P(S|P) = 80\% = 0.80$.
- Probability that a bike from Plant Q is of standard quality, $P(S|Q) = 90\% = 0.90$.
We need to find the probability that the bike came from Plant P, given that it is of standard quality. This is the conditional probability $P(P|S)$.
Calculating Total Probability of Standard Quality
Before applying Bayes' Theorem, we first need to find the total probability that a randomly chosen bike is of standard quality, $P(S)$. A bike can be of standard quality either by coming from Plant P and being standard, or by coming from Plant Q and being standard. We can use the Law of Total Probability:
\( P(S) = P(S|P) \cdot P(P) + P(S|Q) \cdot P(Q) \)
Plugging in the given values:
\( P(S) = (0.80)(0.60) + (0.90)(0.40) \)
\( P(S) = 0.48 + 0.36 \)
\( P(S) = 0.84 \)
So, the total probability of selecting a standard quality bike is 0.84 or 84%.
Applying Bayes' Theorem
Now we can use Bayes' Theorem to find the probability that the bike came from Plant P, given it is of standard quality, $P(P|S)$. Bayes' Theorem is stated as:
\( P(P|S) = \frac{P(S|P) \cdot P(P)}{P(S)} \)
We have all the values needed:
- $P(S|P) = 0.80$
- $P(P) = 0.60$
- $P(S) = 0.84$
Substitute these values into the formula:
\( P(P|S) = \frac{(0.80)(0.60)}{0.84} \)
\( P(P|S) = \frac{0.48}{0.84} \)
To express this as a fraction, we can multiply the numerator and denominator by 100:
\( P(P|S) = \frac{48}{84} \)
Now, simplify the fraction by finding the greatest common divisor (GCD) of 48 and 84. The GCD of 48 and 84 is 12.
Divide the numerator and the denominator by 12:
\( P(P|S) = \frac{48 \div 12}{84 \div 12} = \frac{4}{7} \)
Thus, the probability that the standard quality bike came from Plant P is \( \frac{4}{7} \).
Summary of Calculation
| Probability | Value | Calculation/Source |
|---|---|---|
| P(P) | 0.60 | Given |
| P(Q) | 0.40 | Given |
| P(S|P) | 0.80 | Given |
| P(S|Q) | 0.90 | Given |
| P(S) | 0.84 | \( P(S|P)P(P) + P(S|Q)P(Q) \) |
| P(P|S) | \( \frac{4}{7} \) | \( \frac{P(S|P)P(P)}{P(S)} = \frac{0.48}{0.84} \) |
The final calculated probability is \( \frac{4}{7} \).
Revision Table: Probability Concepts
| Concept | Description | Formula Example |
|---|---|---|
| Conditional Probability | The probability of event A occurring given that event B has already occurred. | \( P(A|B) = \frac{P(A \cap B)}{P(B)} \) |
| Bayes' Theorem | Relates conditional probabilities of two events. Used to update the probability of a hypothesis based on new evidence. | \( P(A|B) = \frac{P(B|A)P(A)}{P(B)} \) |
| Law of Total Probability | Used to find the total probability of an event by considering all possible mutually exclusive cases. | \( P(B) = \sum P(B|A_i)P(A_i) \) (where \( A_i \) partition the sample space) |
| Joint Probability | The probability of two or more events occurring together. | \( P(A \cap B) = P(A|B)P(B) \) or \( P(A \cap B) = P(B|A)P(A) \) |
Additional Information: Probability in Manufacturing
Probability is widely used in manufacturing and quality control. Understanding concepts like conditional probability and Bayes' theorem allows factories to analyze production processes and improve quality.
- Quality Control: By knowing the defect rates from different machines or plants, companies can identify sources of issues.
- Supplier Reliability: If parts come from different suppliers, probability can help determine which supplier is most likely the source of a faulty component.
- Process Improvement: Analyzing conditional probabilities can help pinpoint steps in the manufacturing process that contribute most to defects, allowing for targeted improvements.
- Predictive Maintenance: Probability models can predict the likelihood of machine failure based on operational data, allowing for maintenance before issues occur.
In this problem, using Bayes' theorem helped us update our initial belief about where a bike came from (60% from P, 40% from Q) based on the new information that the bike was of standard quality. The updated probability that it came from Plant P, given it's standard quality (\( \frac{4}{7} \approx 0.571 \)), is slightly lower than the initial probability (0.60). This makes sense because Plant Q has a higher standard quality rate (90%) compared to Plant P (80%), so finding a standard quality bike makes it slightly less likely (though still probable) that it came from Plant P compared to the overall proportion of bikes from P.
Important Questions from Conditional Probability
Let A and B be two events such that \(P(\overline{A \cup B}) = \dfrac{1}{6}\) , P(A ∩ B) = \(\dfrac{1}{4}\) and P(A̅) = \(\dfrac{1}{4}\) , where A̅ stands for complement of event A. Then, events A and B are:
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
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?
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
