Solution:

Let's consider the possible outcomes of the first few throws of the dice:

- If 1 comes in the first throw itself, it will be an even number of trials.
- If 1 comes in the second throw, it will be an odd number of trials.
- If 1 comes in the third throw, it will be an even number of trials.
- If 1 comes in the fourth throw, it will be an odd number of trials.
- If 1 comes in the fifth throw, it will be an even number of trials.
- If 1 comes in the sixth throw, it will be an odd number of trials.
- and so on...

So, we can see that the probability of getting 1 in an even number of throws is the sum of probabilities of getting 1 in the first, third, fifth, seventh, etc. throws.

Let's denote the probability of getting 1 in the first throw as p, and the probability of not getting 1 in the first throw as q. Since the dice is fair, we have:

p = 1/6 and q = 5/6

The probability of getting 1 in the third throw is the probability of not getting 1 in the first two throws (which is q*q) multiplied by the probability of getting 1 in the third throw (which is p):

P(1 on third throw) = q*q*p = (5/6)*(5/6)*(1/6)

Similarly, the probability of getting 1 in the fifth throw is:

P(1 on fifth throw) = q*q*q*q*p = (5/6)*(5/6)*(5/6)*(5/6)*(1/6)

And so on...

So, the probability of getting 1 in an even number of throws is:

P(1 in even number of trials) = P(1 on first throw) + P(1 on third throw) + P(1 on fifth throw) + ...

= p + (q*q*p) + (q*q*q*q*p) + ...

= p * (1 + q*q + q^4 + ...)

= p * (1/(1-q^2))

= (1/6) * (1/(1-(5/6)^2))

= 5/11

Therefore, the correct answer is option A) 5/11.