Question

X speaks truth 4 out of 5 times. A die is tossed and X reports it to be 6. The probability that there was actually 6 is:

• A. \(2/9\)
• B. \(3/9\)
• C. \(4/9\)
• D. \(5/9\)

Hint:

Exam Tip:
For Bayes' theorem problems:

• Define events clearly.
• Compute conditional probabilities carefully.
• Use total probability theorem to find \(P(R)\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This is a problem of conditional probability using Bayes' theorem.
We need to find \(P(\text{Actually 6} \mid \text{Reports 6})\).

Step 2: Key Formula or Approach:
Let \(A\) be the event that the die actually shows 6.
Let \(R\) be the event that X reports 6.
We need \(P(A \mid R)\).
Using Bayes' theorem: \[ P(A \mid R) = \frac{P(R \mid A) P(A)}{P(R)} \]

Step 3: Detailed Explanation:
We have:
• \(P(A) = \frac{1}{6}\) (probability of rolling a 6).
• \(P(\text{not } A) = \frac{5}{6}\).
• \(P(R \mid A) = \frac{4}{5}\) (X speaks truth, so if it is 6, he reports 6).
• \(P(R \mid \text{not } A) = \frac{1}{5}\) (X lies, so he reports 6 when it is not 6).
But wait, if X lies, he could report any number other than the actual.
If the actual is not 6, there are 5 possible outcomes.
The probability that he reports 6 given it is not 6 is \(\frac{1}{5} \times \frac{1}{5} = \frac{1}{25}\)?
Let's think carefully:
If the die shows a number other than 6 (probability \(5/6\)), X lies with probability \(1/5\).
If he lies, he can report any of the 5 numbers other than the actual.
The probability that he reports 6 is \(\frac{1}{5}\) (since there are 5 numbers, and one of them is 6).
So, \(P(R \mid \text{not } A) = \frac{1}{5} \times \frac{1}{5} = \frac{1}{25}\).
This is the correct approach. Alternatively, a simpler approach using the total probability theorem: \[ P(R) = P(R \mid A) P(A) + P(R \mid \text{not } A) P(\text{not } A) \] \[ P(R) = \frac{4}{5} \cdot \frac{1}{6} + \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{5}{6} = \frac{4}{30} + \frac{1}{30} = \frac{5}{30} = \frac{1}{6} \] Wait, that gives \(P(R) = \frac{1}{6}\), which is the same as \(P(A)\).
Then, \[ P(A \mid R) = \frac{\frac{4}{5} \cdot \frac{1}{6}}{\frac{1}{6}} = \frac{4}{5} \] But \(4/5\) is not in the options.
Let's reconsider:
If X lies, he could report any number other than the actual.
If the actual is not 6, there are 5 possible outcomes.
The probability that he reports 6 is \(1/5\).
So, \(P(R \mid \text{not } A) = \frac{1}{5}\) (since he lies with probability \(1/5\) and then chooses 6 with probability \(1/5\)).
Wait, this is incorrect. The probability that he reports 6 given it is not 6 is: \[ P(R \mid \text{not } A) = P(\text{lies}) \times P(\text{reports 6} \mid \text{lies}) = \frac{1}{5} \times \frac{1}{5} = \frac{1}{25} \] Then, \[ P(R) = \frac{4}{5} \cdot \frac{1}{6} + \frac{1}{25} \cdot \frac{5}{6} = \frac{4}{30} + \frac{5}{150} = \frac{20}{150} + \frac{5}{150} = \frac{25}{150} = \frac{1}{6} \] So, \(P(R) = \frac{1}{6}\).
Then, \[ P(A \mid R) = \frac{\frac{4}{5} \cdot \frac{1}{6}}{\frac{1}{6}} = \frac{4}{5} \] But \(4/5\) is not in the options.
Let's check the options: \(2/9, 3/9, 4/9, 5/9\).
There is a possibility that the question assumes that if X lies, he always reports 6.
If X lies and the actual is not 6, he reports 6.
Then, \(P(R \mid \text{not } A) = \frac{1}{5}\).
Then, \[ P(R) = \frac{4}{5} \cdot \frac{1}{6} + \frac{1}{5} \cdot \frac{5}{6} = \frac{4}{30} + \frac{5}{30} = \frac{9}{30} = \frac{3}{10} \] Then, \[ P(A \mid R) = \frac{\frac{4}{5} \cdot \frac{1}{6}}{\frac{3}{10}} = \frac{\frac{4}{30}}{\frac{3}{10}} = \frac{4}{30} \times \frac{10}{3} = \frac{4}{9} \] This gives option (C).
So, the intended interpretation is that if X lies, he reports 6.
Therefore, the probability is \(4/9\).

Step 4: Final Answer:
Therefore, option (C) is correct.