A thermal power plant has an agreement with three mines M1, M2 and M3 to receive 'Grade 1' coal, in the proportion of 60%, 25% and 15%, respectively. The probabilities that a wagon... A thermal power plant has an agreement with three mines M1, M2 and M3 to receive 'Grade 1' coal, in the proportion of 60%, 25% and 15%, respectively. The probabilities that a wagon supplied coal from mines M1, M2 and M3 are 0.02, 0.03 and 0.04, respectively. On a random check, a sample wagon is found to carry below 'Grade 1' coal. The probability that the wagon belongs to mine M1 is _____ (round off up to 2 decimals)
Understand the Problem
The question is asking for the probability that a wagon found to carry 'below Grade 1' coal belongs to mine M1, given certain conditions and probabilities related to coal supply from three mines.
Answer
The probability that the wagon belongs to mine M1 is $0.47$.
Answer for screen readers
The probability that the wagon belongs to mine M1 is $0.47$.
Steps to Solve
- Identify Given Information
Let:
- $P(M1) = 0.6$ (Probability that a wagon is from mine M1)
- $P(M2) = 0.25$ (Probability that a wagon is from mine M2)
- $P(M3) = 0.15$ (Probability that a wagon is from mine M3)
Also, let:
- $P(B|M1) = 0.02$ (Probability that a wagon from mine M1 is below Grade 1)
- $P(B|M2) = 0.03$ (Probability that a wagon from mine M2 is below Grade 1)
- $P(B|M3) = 0.04$ (Probability that a wagon from mine M3 is below Grade 1)
- Calculate Total Probability of Below Grade 1 Coal
Using the law of total probability: $$ P(B) = P(B|M1)P(M1) + P(B|M2)P(M2) + P(B|M3)P(M3) $$ Substituting the values: $$ P(B) = (0.02)(0.6) + (0.03)(0.25) + (0.04)(0.15) $$
- Substituting Values and Simplifying
Calculating each term:
- For M1: $0.02 \times 0.6 = 0.012$
- For M2: $0.03 \times 0.25 = 0.0075$
- For M3: $0.04 \times 0.15 = 0.006$
Combine these results: $$ P(B) = 0.012 + 0.0075 + 0.006 = 0.0255 $$
- Apply Bayes' Theorem
We want to find the probability that the wagon belongs to mine M1 given that it carries below Grade 1 coal: $$ P(M1|B) = \frac{P(B|M1)P(M1)}{P(B)} $$
- Substituting Values into Bayes’ Theorem
Substituting the known values: $$ P(M1|B) = \frac{(0.02)(0.6)}{0.0255} $$ Calculating the numerator: $$ 0.02 \times 0.6 = 0.012 $$ So: $$ P(M1|B) = \frac{0.012}{0.0255} $$
- Calculating Final Value
Now, solving for: $$ P(M1|B) = 0.4706 $$ Rounding to two decimal places, we get: $$ P(M1|B) = 0.47 $$
The probability that the wagon belongs to mine M1 is $0.47$.
More Information
This result is obtained through the application of probability rules and Bayes' theorem, which allows us to update our beliefs based on new evidence.
Tips
- Ignoring the Total Probability: Failing to calculate the total probability of below Grade 1 coal from all mines can lead to incorrect answers. Always ensure to consider all components when using Bayes' theorem.
- Rounding Too Early: It’s a common mistake to round off numbers too soon in calculations. Always keep a few extra decimal places until the final step to maintain accuracy.
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