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I am solving this problem.

Given... $P[W|T]=0.8, \space P[W|T \cap G]=.65, \space P[W | G' \cap T]=1$

Find $P[G'|T]$

I understand that what I am looking for is $$P[G'\cap T]/P[T]$$ or perhaps $$(1-P[G \cap T])/P[T]$$.

I also know that $$P[T]=0.8/P[W\cap T]$$ and since $$P[W\cap T \cap G]/P[T\cap G]=0.65$$, $$P[W\cap T ]=0.65P[T \cap G]/P[W|T \cap G]$$

By the time I am here I am thinking that I am not going in the right direction because I still have unknowns that seems a bit of work to deduce what it is.

Supposedly the answer is $\frac{3}{7}$. Can someone give me an intuitive, or maybe a natural way to solve this problem?

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hyg17
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Define $$\begin{align*} x &= \Pr[W \cap T \cap G], \\ y &= \Pr[T \cap G']. \end{align*}$$ Then since $$\Pr[W \mid (T \cap G')] = \frac{\Pr[W \cap T \cap G']}{\Pr[T \cap G']} = 1,$$ it follows that $$\Pr[W \cap T \cap G'] = \Pr[T \cap G'] = y.$$ So $$\begin{align*} x + y &= \Pr[W \cap T \cap G] + \Pr[W \cap T \cap G'] \\ &= \Pr[W \cap T] = \Pr[W \mid T]\Pr[T] \\ &= \tfrac{4}{5} \Pr[T].\end{align*}$$ Next, consider $$\begin{align*} \frac{13}{20} = \Pr[W \mid (T \cap G)] &= \frac{\Pr[W \cap T \cap G]}{\Pr[T \cap G]} \\ &= \frac{x}{\Pr[T] - \Pr[T \cap G']} \\ &= \frac{x}{\frac{5}{4}(x+y) - y}.\end{align*}$$ Hence $$y = \frac{15}{13}x.$$ It follows that $$\Pr[G' \mid T] = \frac{\Pr[T \cap G']}{\Pr[T]} = \frac{y}{\frac{5}{4}(x+y)} = \frac{3}{7}.$$ This solution is much easier to understand if you draw an appropriate Venn diagram, and consider the relative areas of the regions.

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