Two dice are thrown simultaneously. If X denotes the number of fours, then the expectation of X will be:

Correct option is B

​$$\begin{aligned}&\textbf{Case 1: } X = 0 \text{ (No fours appear)} \\&\text{The probability that neither die shows a four is } \frac{5}{6} \times \frac{5}{6}: \\&\qquad P(X = 0) = \left(\frac{5}{6}\right)^2 = \frac{25}{36} \\\\&\textbf{Case 2: } X = 1 \text{ (Exactly one four appears)} \\&\qquad P(X = 1) = 2 \times \left(\frac{1}{6} \times \frac{5}{6}\right) = 2 \times \frac{5}{36} = \frac{10}{36} \\\\&\textbf{Case 3: } X = 2 \text{ (Exactly two fours appear)} \\&\text{The probability that both dice show a four is } \frac{1}{6} \times \frac{1}{6}: \\&\qquad P(X = 2) = \frac{1}{36} \\\\&\textbf{Step 3: Calculate the expected value } E(X) \\&\text{The expected value of } X, \text{ denoted } E(X), \text{ is given by the sum of the products of each value of } X \text{ and its} \\&\text{corresponding probability:} \\&\qquad E(X) = 0 \times P(X = 0) + 1 \times P(X = 1) + 2 \times P(X = 2) \\&\text{Substituting the probabilities:} \\&\qquad E(X) = 0 \times \frac{25}{36} + 1 \times \frac{10}{36} + 2 \times \frac{1}{36} \\&\qquad E(X) = 0 + \frac{10}{36} + \frac{2}{36} \\&\qquad E(X) = \frac{12}{36} = \frac{1}{3}\end{aligned}$$​​