Probability of Even Score with 2 Tetrahedral Dice

[scolty]

Hi,

im trying to figure out how to solve the following question and any hints would be greatly appreciated.

Question: Two tetrahedral dice, with faces labelled 1,2,3,4 are thrown and the number on which each lands is noted. The score is the sum of the two faces. Find the probability that the score is even, given that at least one die lands on a 3.

The final answer is suppose to be 3/7.

Below is as far as i had managed to get.

P(A|B) = P(A&B)/P(B)

where:
A is the sum = an even number, ie 2nd dice = 1 or 3
B the first dice is 3.

if the above is correct (ie P(B) isn't the probability that a single dice is showing 3) then is 1/4, otherwise i think its 3/16 (that may be wrong.)

im lost from here on. As i mentioned earlier, any assistance would be appreciated.

 

[scolty]

i managed to solve this by writing out the sequence of dice rolls as follows:
(sure 99% of the ppl using this forum know more about stats than i do, so this is for the 1%)

green are the possible scenarios where one die is a 3.

red are the corresponding scenarios where the total is an even number

P(A|B) = P(A&B) / P(B)

where P(B) = 7/16 (note I am not entirely sure why its not 8 since 3,3 isn't counted as two possibilities.)
P(A&B) = 3/16 (See the combinations which are both green and red)

answer = 3/7