Question

Consider a case where two identical six-faced fair dice are rolled simultaneously.

The probability of getting even numbers in both AND a sum equal to 5 and above is ______ (rounded off to two decimal places).

Hint:

First list the outcomes where both dice are even, then keep only those with sum 5 or more.

Answer 1

Correct Answer: 0.22

Step 1: Separate the two conditions.
Let event $E$ be "both dice show even numbers" and event $S$ be "the sum is 5 or more." We need $P(E \cap S)$.

Step 2: Find $P(E)$ and then restrict by $S$.
$P(E) = \dfrac{3}{6}\times\dfrac{3}{6} = \dfrac{9}{36}$, since each die has 3 even faces out of 6.
Within these 9 even-even pairs, only the pair (2,2) gives a sum below 5, since $2+2=4$. Every other even-even pair sums to 6 or more.

Step 3: Remove the failing pair and convert to probability.
So $9 - 1 = 8$ of the 9 even-even pairs also satisfy the sum condition, giving 8 favorable outcomes out of the full 36 outcomes.
$P(E \cap S) = \dfrac{8}{36} = 0.2222...$, which rounds to $0.22$.

Final Answer:
The required probability rounds to 0.22.
\[ \boxed{0.22} \]