Two identical cube shaped dice each with faces numbered $1$ to $6$ are rolled simultaneously. The probability that an even number is rolled out on each dice is:

1. $\frac{1}{36}$
2. $\frac{1}{12}$
3. $\frac{1}{8}$
4. $\frac{1}{4}$

Two identical cube-shaped dice each with faces numbered $1$ to $6$ are rolled simultaneously. Then all possible outcomes.

• $(1,1), (1,2),(1,3),(1,4),(1,5),(1,6)$
• $(2,1), {\color{Red} {(2,2)}},(2,3),{\color{Red} {(2,4)}},(2,5),{\color{Red} {(2,4)}}$
• $(3,1), (3,2),(3,3),(3,4),(3,5),(3,6)$
• $(4,1), {\color{Blue} {(4,2)}},(4,3),{\color{Blue} {(4,4)}},(4,5),{\color{Blue} {(4,6)}}$
• $(5,1), (5,2),(5,3),(5,4),(5,5),(5,6)$
• $(6,1), {\color{Green} {(6,2)}},(6,3),{\color{Green} {(6,4)}},(6,5),{\color{Green} {(6,6)}}$

The probability that an even number is rolled out on each dice is $= \dfrac{9}{36} = \dfrac{1}{4}.$

$\textbf{Short Method:}$ Two identical cube-shaped dice each with faces numbered $1$ to $6$ are rolled simultaneously.

Dice $1: 1,{\color{Magenta} {2}},3,{\color{Magenta} {4}},5,{\color{Magenta} {6}}$

Dice $2: 1,{\color{Teal} {2}},3,{\color{Teal} {4}},5,{\color{Teal} {6}}$

• $P(\text{Even number on first dice)} = P(E_{1}) = \dfrac{3}{6} = \dfrac{1}{2}$
• $P(\text{Even number on second dice)} = P(E_{2}) = \dfrac{3}{6} = \dfrac{1}{2}$

Now, $P(\text{Even number on both dice}) = P(E_{1} \cap E_{2}) = P(E_{1}) \cdot P(E_{2}) = \dfrac{1}{2} \cdot \dfrac{1}{2} = \dfrac{1}{4}.$

$\textbf{PS:}$ These two events are independent events.

So, the correct answer is $(D).$

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