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 that occurs together with any one of E₁, E₂, ..., or E_n, such that P(A) ≠ 0.

Then, according to Bayes' theorem:

\[\rm P\left( E_i\;|\;A \right) = \frac{P\left( E_i \right) \times P\left( A\;|\;E_i \right)}{\sum\limits_{i = 1}^{n} P\left( E_i \right) \times P\left( A\;|\;E_i \right)},\quad i = 1,2,\dots,n\]

Calculation:

Let:

E₁: He knows the answer

E₂: He does not know the answer

X: He gets the correct answer.

Given:

P(E₁) = 90% = \(\frac{9}{10}\)

P(E₂) = \(1 - \frac{9}{10} = \frac{1}{10}\)

P(X | E₁) = 1

P(X | E₂) = \(\frac{1}{4}\)

Using Bayes' theorem:

\[\rm P(E_2\;|\;X) = \frac{P(E_2) \cdot P(X\;|\;E_2)}{P(E_1) \cdot P(X\;|\;E_1) + P(E_2) \cdot P(X\;|\;E_2)}\]

\[\rm P(E_2\;|\;X) = \frac{\frac{1}{10} \times \frac{1}{4}}{\left( \frac{9}{10} \times 1 + \frac{1}{10} \times \frac{1}{4} \right)} = \frac{1}{37}\]