2 four sided dice, probability of 1 dice = 3, sum = even

[Question] 2 four tetrahedral dice with sides, 1,2,3,4. the two dice are thrown, what is the probability that given the first dice is a 3, the sum of the 2 dices will be an even number.

This has had me stumped for about 30mins. im assuming i should be using

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

P(B) = Rolling a single 3 from both dices? or should this just be rolling a 3 on a single dice.

P(A) = Sum of the numbers = an even number, of which there are 2 option: 3+1 and 3+3

Answer is supposedly 3/7

If someone could give me some pointers on this id appreciate it.

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 (Edit: Its only 7 and not 8 as B is the case where only one 3 is encountered)
P(A&B) = 3/16 (See the combinations which are both green and red)

answer = 3/7
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