A speaks the truth 4 out of 5 times. He throws a die and reports that there was a 6, the probability that actually there was a 6 is

Options

A. `4//9`
B. `5//9`
C. `3//10`
D. none of these

Answer

To solve the problem, we will use Bayes' theorem. Let's break it down step by step.

### Step 1: Define the Events
Let:
- \( A \): the event that the die actually shows a 6.
- \( B \): the event that A reports that the die shows a 6.

### Step 2: Determine the Probabilities
From the problem statement:
- The probability that A speaks the truth (reports correctly) is \( P(B|A) = \frac{4}{5} \).
- The probability that A does not speak the truth (reports incorrectly) is \( P(B|A') = \frac{1}{5} \) (where \( A' \) is the event that the die does not show a 6).
- The probability that the die shows a 6 is \( P(A) = \frac{1}{6} \).
- The probability that the die does not show a 6 is \( P(A') = 1 - P(A) = 1 - \frac{1}{6} = \frac{5}{6} \).

### Step 3: Calculate \( P(B) \)
Using the law of total probability, we can calculate \( P(B) \):
\[ P(B) = P(B|A) \cdot P(A) + P(B|A') \cdot P(A') \]
Substituting the values we have:
\[ P(B) = \left(\frac{4}{5} \cdot \frac{1}{6}\right) + \left(\frac{1}{5} \cdot \frac{5}{6}\right) \]
Calculating each term:
- First term: \( \frac{4}{5} \cdot \frac{1}{6} = \frac{4}{30} \)
- Second term: \( \frac{1}{5} \cdot \frac{5}{6} = \frac{5}{30} \)

Now adding these:
\[ P(B) = \frac{4}{30} + \frac{5}{30} = \frac{9}{30} = \frac{3}{10} \]

### Step 4: Apply Bayes' Theorem
Now, we want to find \( P(A|B) \), the probability that the die actually shows a 6 given that A reports it shows a 6. According to Bayes' theorem:
\[ P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)} \]
Substituting the values:
\[ P(A|B) = \frac{\left(\frac{4}{5}\right) \cdot \left(\frac{1}{6}\right)}{\frac{3}{10}} \]
Calculating the numerator:
\[ \frac{4}{5} \cdot \frac{1}{6} = \frac{4}{30} \]
Now substituting into Bayes' theorem:
\[ P(A|B) = \frac{\frac{4}{30}}{\frac{3}{10}} = \frac{4}{30} \cdot \frac{10}{3} = \frac{4 \cdot 10}{30 \cdot 3} = \frac{40}{90} = \frac{4}{9} \]

### Final Answer
The probability that the die actually shows a 6 given that A reports it shows a 6 is \( \frac{4}{9} \).

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