How to find $P(B \mid T) $ and $P(B \mid T^c)$ given $P(T \mid B)$ and $P(T \mid B^ c)$

Question

How can one calculate $P(B\mid T) $ and $P(B\mid T^c)$ given $P(T \mid B)=0.75$ and $P(T \mid B^ c)=0.15$ ?

I have found the answers. However, I have solved it in seven pages. I am getting $P(B|T)=0.75$ and $P(B|T^c)=0.09375$.

Would someone supply a better and shorter answer. Thanks in anticipation.

Answer 1

The answer is not correct, I think. And I also think the information is insufficient.

Setup

Let us call $$P(B \cap T)=a$$ $$P(B - (B \cap T))=b$$ $$P(T - (B \cap T))=c$$ $$P((B \cup T)^c)=d$$

Information we have

$$a + b + c + d = 1\tag{1}\label{1}$$ $$\frac{a}{a + b} = 0.75\tag{2}\label{2}$$ $$\frac{c}{c + d} = 0.15\tag{3}\label{3}$$

Goal

The target is to calculate $a/(a + c)$ and $b/(b + d)$.

From the fact that we just have three equations in 4 variables, it seems that it is not possible to get all the four variables. But it might still be possible to get the two terms we want, so I have tried to show below that that is not the case.

Algebraic rearrangements

From $(2)$ we have $a = 3b$

From $(3)$ we have $c = 3d/17$

Substituting these values in $(1)$ gives us $$4b + \frac{20d}{17} = 1$$ or equivalently $$d = \frac{(1 - 4b)17}{20}\tag{4}\label{4}$$ Using $(4)$ we can get the two desired equations in the terms of $b$. $$\frac{a}{a + c} = \frac{3b}{3b + \frac{3(1 - 4b)}{20}}$$ $$\frac{b}{b + d} = \frac{b}{b + \frac{(1-4b)17}{20}}$$

Since there is no way to further reduce this set of equations, we cannot get the value of these two terms without knowing $b$.

You can get as many answers as you want by substituting different values of $b$. Furthermore, the two answers you give, are not compatible with one another (i.e. do not fit into this equation). Another way to see the incompatibility is as follows:

$$P(T|B)P(B) = P(B|T)P(T)$$ Since $P(T|B)=P(B|T)$, we have $P(B)=P(T)=x$.

Next, using $P(T^c|B)=1 - P(T|B) = 0.25$ and $$P(T^c|B)x = P(B|T^c)(1 - x)$$ which gives us $$0.25x = 0.09375(1 - x)$$ But, if we use $P(B^c|T) = 1 - P(B|T) = 0.25$ and $$P(T|B^c)(1 -x)=P(B^c|T)x$$ we get $$0.15(1 - x) = 0.25x$$ These two are clearly incompatible.