Step-by-step Solution:

Concept:

Bayes' Theorem:

Let E₁, E₂, …, E_n be n mutually exclusive and exhaustive events associated with a random experiment and let S be the sample space. Let A be any event which occurs together with any one of E₁ or E₂ or … or E_n such that P(A) ≠ 0. Then

\(P\left( {{E_i}\;|\;A} \right) = \frac{{P\left( {{E_i}} \right)\; \times \;P\left( {A\;|\,{E_i}} \right)}}{{\sum_{i = 1}^n P\left( {{E_i}} \right) \;\times\; P\left( {A\;|\,{E_i}} \right)}}\), \(i = 1,\;2,\; \ldots .\;,n\)

Calculations:

Given, A man is known to speak the truth 2 out of 3 times.

E₁ : Probability that man speaks the truth = P(E₁) = \( \dfrac{2}{3} \)
E₂ : The probability that man lies = P(E₂) = \( \dfrac{1}{3} \)

P (X | E₁) : Probability of getting 1 on die face = \( \dfrac{1}{6} \)

P (X | E₁) : Probability of not getting 1 on die face = \( \dfrac{5}{6} \)

Here, we have to find the value of P(E₁ | X).

As we know that according to Bayes' theorem: \( P\left( {{E_i}\;|\;A} \right) = \frac{{P\left( {{E_i}} \right)\; \times \;P\left( {A\;|\,{E_i}} \right)}}{{\sum_{i = 1}^n P\left( {{E_i}} \right)\; \times\; P\left( {A\;|\,{E_i}} \right)}} \)

\( \Rightarrow P(E_1 | X) = \frac{{P\left( {{E_1}} \right) \cdot P\left( {X | {E_1}} \right)}}{{\left[ {P\left( {{E_1}} \right) \cdot P\left( {X | {E_1}} \right) + P\left( {{E_2}} \right) \cdot P\left( {X | {E_2}} \right)} \right]}} \)

\( \Rightarrow P(E_1 | X) = \frac{{\frac{1}{6} \cdot \frac{2}{3}}}{{\left[ {\frac{1}{6} \cdot \frac{2}{3} + \frac{5}{6} \cdot \frac{1}{3}} \right]}} = \frac{2}{7} \)