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My problem statement is as follows: If $$P(Y=0 | X=0) = 1-\xi_{1} $$ and $$P(Y=1 | X=1) = 1-\xi_{2} $$, what is the probability of $P(Y=0)$?

I am trying to set this up starting with the formula for conditional probability, but I am getting stuck in trying to extract $P(Y=0)$ by itself.

Thanks for your responses :)

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    we need a lot more information than you've given to solve this problem. – 2017-02-01
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    By the Law of Total Probability $$\mathsf P(Y=0)~=~\underbrace{\mathsf P(Y=0\mid X=0)}_{1-\xi_1}~\mathsf P(X=0)+\mathsf P(Y=0\mid X\neq 0)~\mathsf P(X\neq 0)$$ You have only provided one factor on the RHS.   Without means to evaluate the other three, we cannot proceed. – 2017-02-01
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    For instance, I assume the only values $X$ and $Y$ can take are $0$ and $1$? If so you can say $P(Y=0) = P(Y=0|X=0)P(X=0)+P(Y=0|X=1)P(X=1)$ but you need to know $P(X=0)$ and $P(X=1).$ – 2017-02-01
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    @GrahamKemp How did you use the Law of Total Probability to arrive at your answer? Could you write it as an answer, so I can accept? – 2017-02-02

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By the Law of Total Probability $\mathsf P(A)=\mathsf P(A\mid B)~\mathsf P(B)+\mathsf P(A\mid B^\complement)~\mathsf P(B^\complement)$, then:

$$\begin{align}\\ \mathsf P(Y=0)~&=~\mathsf P(Y=0\mid X=0)~\mathsf P(X=0)+\mathsf P(Y=0\mid X\neq 0)~\mathsf P(X\neq 0) \\[2ex] &=~(1-\xi_1)~\mathsf P(X=0)+\mathsf P(Y=0\mid X\neq 0)~\mathsf P(X\neq 0)\\ \;\end{align}$$

You have only provided one factor on the RHS.   Without means to evaluate the other three, we cannot proceed.

For instance, if we can assume the only values $X$ and $Y$ may take are $0$ and $1$, (may we?) then if so we can say

$$\begin{align}\mathsf P(Y=0)~&=~\mathsf P(Y=0\mid X=0)~\mathsf P(X=0)+\big(1-\mathsf P(Y=1\mid X=1)\big)~\mathsf P(X=1)\\[2ex] &=~(1-\xi_1)~\mathsf P(X=0)+\xi_2~\mathsf P(X= 1)\end{align}$$

However, we will still need to know $\mathsf P(X=0)$ and $\mathsf P(X=1)$.