Given that two dice are thrown, and the first is an even number, what is the conditional probability of getting two sixes?
I am unsure of whether to calculate $P(\text{two}~6~\text{'s} \mid \text{even number})$ as $1/6$ multiplied by $1/6$, and divided by $1/2$, or as $1/2$ (since the first number is an even number) multiplied by $1/6$, divided by $1/2$ (the first of my attempts is the correct answer, $1/18$).
Find the conditional probability of getting at least one six given that the first number is an even number.
I assumed this was $1-(5/6)^2$, divided by $1/2$? The answer is $4/9$.
Independence of the dice rolls, so $$P(x=6,y=6|x\in\{2,4,6\}) = P(x=6|x\in\{2,4,6\})\cdot P(y=6)=1/3 * 1/6$$
For your second question, $1-(5/6)^2$ is the probability of getting at least one six, not the conditional probability given the first is even. (It’s not 5/6 ways to not get a six if you already know its even.)
As for dividing by 2, this doesnt work because when you get at least one six, the first one isnt even half the time and odd half the time. The reason is, sometimes only the first one is a six, and in that case the first one must be even (not 50% odd 50% even).