Odds in favour of A's speaking the thuth are 1: 2.
odds against B's speaking the truth Are 1: 3
 A die is thrown. Both A and B assert that on the die 4 has turned up. Then the probability of the truth of their assertion is ?

To solve this problem, we use conditional probability along with the given odds.

### Step 1: Understand the problem

- **Odds in favor of A speaking the truth** are 1:2. This means the probability of A telling the truth is:
  \[
  P(A \text{ tells the truth}) = \frac{1}{1+2} = \frac{1}{3}
  \]
  and the probability of A lying is:
  \[
  P(A \text{ lies}) = 1 - \frac{1}{3} = \frac{2}{3}
  \]

- **Odds against B speaking the truth** are 1:3. This means the probability of B telling the truth is:
  \[
  P(B \text{ tells the truth}) = \frac{3}{3+1} = \frac{3}{4}
  \]
  and the probability of B lying is:
  \[
  P(B \text{ lies}) = 1 - \frac{3}{4} = \frac{1}{4}
  \]

### Step 2: Approach
The event we're interested in is whether both A and B are telling the truth, given that they both assert the die shows 4.

Define the following events:
- Let \( T \) be the event that the die actually shows 4.
- Let \( A_T \) be the event that A says the die shows 4.
- Let \( B_T \) be the event that B says the die shows 4.

We need to calculate \( P(T \mid A_T \cap B_T) \), the probability that the die actually shows 4 given that both A and B claim it shows 4.

Using **Bayes' Theorem**:
\[
P(T \mid A_T \cap B_T) = \frac{P(T) \cdot P(A_T \cap B_T \mid T)}{P(A_T \cap B_T)}
\]
Here:
- \( P(T) = \frac{1}{6} \), since a fair die has 6 faces.
- \( P(A_T \cap B_T \mid T) = P(A \text{ tells truth}) \cdot P(B \text{ tells truth}) = \frac{1}{3} \times \frac{3}{4} = \frac{1}{4} \).

Now, we need to find \( P(A_T \cap B_T) \). This is the probability that both A and B assert the die shows 4. This can happen in two mutually exclusive ways:
1. Both A and B tell the truth, and the die shows 4.
2. Either or both lie about the die showing 4.

Thus:
\[
P(A_T \cap B_T) = P(T) \cdot P(A \text{ tells truth}) \cdot P(B \text{ tells truth}) + P(T^c) \cdot P(A \text{ lies}) \cdot P(B \text{ lies})
\]
Where \( T^c \) is the event that the die does not show 4, and \( P(T^c) = \frac{5}{6} \).

So, we have:
\[
P(A_T \cap B_T) = \frac{1}{6} \times \frac{1}{3} \times \frac{3}{4} + \frac{5}{6} \times \frac{2}{3} \times \frac{1}{4}
\]
Simplifying:
\[
P(A_T \cap B_T) = \frac{1}{24} + \frac{10}{72} = \frac{1}{24} + \frac{5}{36} = \frac{1}{24} + \frac{10}{72} = \frac=

Odds in favour of A's speaking the truth are 1:2. Odds against B's speaking the truth are 1:3. A die is thrown. Both A and B assert that on the die, 4 has turned up. What is the probability of the truth of their assertion?

Solve a probability problem using conditional probability and Bayes' Theorem. Given the odds of A and B speaking the truth, find the probability that both are correct when asserting a die roll shows a 4. This solution involves key concepts like conditional probability and Bayes' Theorem.

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