Let a, b, c ∈ \ 1, 2, 3, 4\ . If the probability, that ax^2 + 2 2...

For the quadratic expression $ax^2 + 2\sqrt{2}bx + c > 0$ for all $x \in \mathbb{R}$, the leading coefficient must be positive and the discriminant must be negative.

Since $a \in \{1, 2, 3, 4\}$, the condition $a > 0$ is always satisfied.

The discriminant condition is:
$D < 0$
$\Rightarrow (2\sqrt{2}b)^2 - 4ac < 0$
$\Rightarrow 8b^2 - 4ac < 0$
$\Rightarrow 2b^2 < ac$

The total number of possible triplets $(a, b, c)$ is $4 \times 4 \times 4 = 64$.

Now, we find the number of favorable outcomes by checking the possible values of $b \in \{1, 2, 3, 4\}$:

Case 1: $b = 1$
We need $2(1)^2 < ac \Rightarrow ac > 2$.
The total number of pairs $(a, c)$ is $4 \times 4 = 16$.
The pairs for which $ac \le 2$ are $(1, 1), (1, 2), (2, 1)$, which are $3$ in number.
So, the number of pairs with $ac > 2$ is $16 - 3 = 13$.

Case 2: $b = 2$
We need $2(2)^2 < ac \Rightarrow ac > 8$.
The possible pairs $(a, c)$ from the given set are $(3, 3), (3, 4), (4, 3), (4, 4)$.
So, there are $4$ pairs.

Case 3: $b = 3$
We need $2(3)^2 < ac \Rightarrow ac > 18$.
Since the maximum possible value of $ac$ is $4 \times 4 = 16$, there are $0$ pairs.

Case 4: $b = 4$
We need $2(4)^2 < ac \Rightarrow ac > 32$.
Again, there are $0$ pairs.

Total number of favorable outcomes = $13 + 4 = 17$.

The required probability is $\dfrac{m}{n} = \dfrac{17}{64}$.

Since $\gcd(17, 64) = 1$, we have $m = 17$ and $n = 64$.

Therefore, $m + n = 17 + 64 = 81$.

Answer: $81$