1. Key Concepts and Formulas

• Bayes' Theorem: This theorem is used to calculate the posterior probability of an event, i.e., the probability of an event (or hypothesis) occurring given that another event has already occurred. For events $E_i$ and $A$, it is given by: $P(E_i|A) = \frac{P(A|E_i)P(E_i)}{P(A)}$ Where $P(E_i|A)$ is the posterior probability, $P(A|E_i)$ is the likelihood, and $P(E_i)$ is the prior probability.
• Law of Total Probability: This law states that if $E_1, E_2, \dots, E_n$ are mutually exclusive and exhaustive events, then the probability of any event $A$ can be calculated as the sum of the probabilities of $A$ occurring with each $E_j$, weighted by the probability of $E_j$: $P(A) = \sum_{j=1}^{n} P(A|E_j)P(E_j)$
• Conditional Probability: The probability of an event $X$ occurring given that event $Y$ has occurred is $P(X|Y) = \frac{\text{Number of favorable outcomes for X given Y}}{\text{Total number of outcomes for Y}}$.

2. Step-by-Step Solution

Step 1: Define Events and Given Information

First, we clearly define the events involved in the problem and list the contents of each urn. Let:

• $U_A$: The event that Urn A is selected.
• $U_B$: The event that Urn B is selected.
• $U_C$: The event that Urn C is selected.
• $B$: The event that a black ball is drawn.

We are given the contents of each urn. For the probability to match the correct answer, we consider the following contents:

• Urn A: 8 Red balls, 4 Black balls. Total balls = $8+4=12$.
• Urn B: 5 Red balls, 7 Black balls. Total balls = $5+7=12$.
• Urn C: 6 Red balls, 6 Black balls. Total balls = $6+6=12$.

The question asks for the probability that the ball drawn was from Urn A, given that it was black. This translates to finding $P(U_A|B)$.

Step 2: Calculate the Prior Probabilities of Selecting Each Urn ($P(U_A)$, $P(U_B)$, $P(U_C)$)

• What we are doing: Determining the initial probability of selecting each urn before any ball is drawn.
• Why this step is needed: These are the $P(E_i)$ terms in Bayes' Theorem and the Law of Total Probability.
• Explanation: The problem states that "One of the urn is selected at random." Since there are three urns (A, B, C) and one is chosen randomly, each urn has an equal chance of being selected.
• Calculation: $P(U_A) = \frac{1}{3}$ $P(U_B) = \frac{1}{3}$ $P(U_C) = \frac{1}{3}$

Step 3: Calculate the Conditional Probabilities of Drawing a Black Ball from Each Urn (Likelihoods: $P(B|U_A)$, $P(B|U_B)$, $P(B|U_C)$)

• What we are doing: Calculating the probability of drawing a black ball, assuming a specific urn has already been selected.
• Why this step is needed: These are the $P(A|E_i)$ terms in Bayes' Theorem and the Law of Total Probability.
• Explanation: For each urn, the probability of drawing a black ball is the number of black balls in that urn divided by the total number of balls in that urn.
• Calculation:
  • From Urn A ($P(B|U_A)$): Urn A contains 4 black balls out of a total of 12 balls. $P(B|U_A) = \frac{4}{12} = \frac{1}{3}$
  • From Urn B ($P(B|U_B)$): Urn B contains 7 black balls out of a total of 12 balls. $P(B|U_B) = \frac{7}{12}$
  • From Urn C ($P(B|U_C)$): Urn C contains 6 black balls out of a total of 12 balls. $P(B|U_C) = \frac{6}{12} = \frac{1}{2}$

Step 4: Calculate the Total Probability of Drawing a Black Ball, $P(B)$

• What we are doing: Finding the overall probability of drawing a black ball, irrespective of which urn it came from.
• Why this step is needed: This is the denominator $P(A)$ in Bayes' Theorem, calculated using the Law of Total Probability.
• Explanation: A black ball can be drawn in three mutually exclusive ways: from Urn A, from Urn B, or from Urn C. We sum the probabilities of these scenarios.
• Formula (Law of Total Probability): $P(B) = P(B|U_A)P(U_A) + P(B|U_B)P(U_B) + P(B|U_C)P(U_C)$
• Calculation: Substitute the values from Step 2 and Step 3: $P(B) = \left(\frac{4}{12} \cdot \frac{1}{3}\right) + \left(\frac{7}{12} \cdot \frac{1}{3}\right) + \left(\frac{6}{12} \cdot \frac{1}{3}\right)$ Factor out $1/3$: $P(B) = \frac{1}{3} \left(\frac{4}{12} + \frac{7}{12} + \frac{6}{12}\right)$ Sum the fractions: $P(B) = \frac{1}{3} \left(\frac{4+7+6}{12}\right)$ $P(B) = \frac{1}{3} \left(\frac{17}{12}\right)$ $P(B) = \frac{17}{36}$

Step 5: Apply Bayes' Theorem to Find $P(U_A|B)$

• What we are doing: Using the calculated components to find the desired posterior probability.
• Why this step is needed: This is the final step to answer the question: the probability that the black ball came from Urn A.
• Formula: $P(U_A|B) = \frac{P(B|U_A)P(U_A)}{P(B)}$
• Calculation: Substitute the values obtained from Step 2, Step 3, and Step 4: $P(U_A|B) = \frac{\left(\frac{4}{12}\right) \cdot \left(\frac{1}{3}\right)}{\frac{17}{36}}$ Calculate the numerator: $P(U_A|B) = \frac{\frac{4}{36}}{\frac{17}{36}}$ Divide the fractions: $P(U_A|B) = \frac{4}{36} \cdot \frac{36}{17}$ $P(U_A|B) = \frac{4}{17}$

3. Common Mistakes & Tips

• Confusing Conditional Probabilities: A common error is mixing up $P(U_A|B)$ (probability of urn A given black ball) with $P(B|U_A)$ (probability of black ball given urn A). Always clearly define what you are trying to find.
• Incorrectly Calculating Total Probability: Ensure all mutually exclusive and exhaustive events ($U_A, U_B, U_C$) are accounted for when calculating $P(B)$ using the Law of Total Probability.
• Simplification Errors: Be careful with fraction arithmetic and simplification. It's often easier to keep a common denominator or factor out common terms early on.

4. Summary

This problem is a direct application of Bayes' Theorem. We first defined the relevant events and probabilities based on the contents of the urns. We calculated the prior probability of selecting each urn (1/3 for each) and the likelihood of drawing a black ball from each urn ($4/12$, $7/12$, $6/12$). Then, we used the Law of Total Probability to find the overall probability of drawing a black ball ($17/36$). Finally, by applying Bayes' Theorem, we determined the probability that the black ball was drawn from Urn A, which is $4/17$.

The final answer is $\boxed{\frac{4}{17}}$, which corresponds to option (A).