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

Faster method using linearity of expectation:

For each die, the expected number of fours is $\dfrac{1}{6}$.

Since we have two independent dice:

$E(X) = E(X_1) + E(X_2) = \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{2}{6} = \dfrac{1}{3}$

Therefore, the expectation of $X$ is $\dfrac{1}{3}$.

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Alternate Method:

Let $X$ denote the number of fours that appear when two dice are thrown.

We need to find the expected value $E(X)$.

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Possible values of $X$ are 0, 1, and 2.

When rolling a single die:

• Probability of rolling a 4 = $\dfrac{1}{6}$
  • Probability of not rolling a 4 = $\dfrac{5}{6}$

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Calculating the probability for each value of $X$:

For $X = 0$ (no fours appear):

$P(X = 0) = \dfrac{5}{6} \times \dfrac{5}{6} = \dfrac{25}{36}$

For $X = 1$ (exactly one four appears):

$P(X = 1) = \left(\dfrac{1}{6} \times \dfrac{5}{6}\right) + \left(\dfrac{5}{6} \times \dfrac{1}{6}\right) = \dfrac{10}{36}$

For $X = 2$ (both dice show four):

$P(X = 2) = \dfrac{1}{6} \times \dfrac{1}{6} = \dfrac{1}{36}$

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Calculating the expectation using $E(X) = \sum x \times P(X = x)$:

$E(X) = 0 \times P(X = 0) + 1 \times P(X = 1) + 2 \times P(X = 2)$

$E(X) = 0 \times \dfrac{25}{36} + 1 \times \dfrac{10}{36} + 2 \times \dfrac{1}{36}$

$E(X) = \dfrac{10}{36} + \dfrac{2}{36} = \dfrac{12}{36} = \dfrac{1}{3}$