Finding $P(y|x)$ given $P(y), P(x|y)$, and $P(\lnot x|\lnot y)$.

Question

I have a probability question that asks me to find $P(y|x)$ given that $P(y) = 0.6, P(x|y) = 0.8$, and $P(\lnot x|\lnot y) = 0.3$.

I've tried applying Bayes's theorem and using it I can find $P(x\wedge y) = 0.48$ and $P(\lnot x\wedge \lnot y) = 0.12$. The roadblock I run into is that I seem to need to know $P(x)$ to find it. Any help would be greatly appreciated.

Answer 1

From $P(x\mid \lnot y)+P(\lnot x\mid\lnot y)=1$ find $P(x\mid \lnot y)$. Then from $P(x\mid \lnot y)=\frac{P(x, \lnot y)}{P(\lnot y)}$ find $P(x, \lnot y)$. Finally, $P(x, \lnot y) + P(x, y) = P(x)$.

Answer 2

\begin{align*} \mathbb{P}(Y|X) = \frac{\mathbb{P}(Y\cap X)}{\mathbb{P}(X)} & = \frac{\mathbb{P}(Y)\mathbb{P}(X|Y)}{\mathbb{P}(X\cap Y) + \mathbb{P}(X\cap Y^{c})}\\\\ & = \frac{\mathbb{P}(Y)\mathbb{P}(X|Y)}{\mathbb{P}(Y)\mathbb{P}(X|Y) + \mathbb{P}(Y^{c}) - \mathbb{P}(X^{c}\cap Y^{c})}\\\\ & = \frac{\mathbb{P}(Y)\mathbb{P}(X|Y)}{\mathbb{P}(Y)\mathbb{P}(X|Y) + \mathbb{P}(Y^{c}) - \mathbb{P}(Y^{c})\mathbb{P}(X^{c}|Y^{c})}\\\\ & = \frac{\mathbb{P}(Y)\mathbb{P}(X|Y)}{\mathbb{P}(Y)\mathbb{P}(X|Y) + \mathbb{P}(Y^{c})(1 - \mathbb{P}(X^{c}|Y^{c}))}\\\\ & = \frac{0.6\times 0.8}{0.6\times 0.8 + (1 - 0.6)\times(1 - 0.3)} \end{align*}