If two fair dice are rolled, what is the conditional probability that the first one lands on 6 given that the sum of the dice is i? compute for all values of i between 2 and 12.

So we need to know the likelihood for each sum. the first sum is 2, and there is no way for one of the die to equal 6 if the sum is 2, therefore the probability is 0. The same applies when the sum is 3, 4, 5, and 6. Once the sum gets to 7, you must evaluate all possible options.
For 7, your options are 1&6, 2&5, 3&4, 4&3, 5&2, 6&1, where the number before the ampersand is the first die, and the number after is the second. there is only one option of the 6 choices where the first die is 6, therefore the probability is 1/6.
For 8, the options are 2&6, 3&5, 4&4, 5&3, 6&2. so of the 5 choices, there is only one option, therefore the probability is 1/5.
For 9, the choices are 3&6, 4&5, 5&4, 6&3. So of the 4 choices, there is 1 option, therefore the probability is 1/4. 
For 10, the options are 4&6, 5&5, 6&4. Of the 3 choices, there is 1 option, therefore the probability is 1/3.
For 11, the options are 5&6, 6&5. Of the 2 choices, there is 1 option, therefore the probability is 1/2.
Finally, for 12, the only option is 6&6. There is only one choice, so the probability is 1.

First step is to get rid of the square root. remember that a square root is the same as raising to the (1/2) power. see below: multiply exponents per exponent rules to get: remember you subtract exponents from the denominator and add exponents in the numerator so: simplify further: so a is the correct choice. if you don't understand this google exponent rules and it will you understand. look at the image results.