From the given conditional probability, find P(B)

Question

I'm studying for my exam and came across to this one question that I could not solve. Given $P(A|B)=0.4=P(A'|B')$ and $P(A)=0.5$, find $P(B)$. The answer is $0.5$.

From above, I find that

$0.6P(B)=P(A'\cap B)$

$0.6P(B')=P(A\cap B')$

However when I tried to solve for $P(B)$, everything seems to circulate back to the original clue given.

Answer 1

Here is the logs of an experiment made hundreds of years ago.

Unfortunately the middle is missing. Unfortunately because

we need the number of $1$s in the $B$ column.

Luckily some written notes remained. We still know that in the $A$ column $40\%$ of the yellow rows contain $1$s, and also, in the $A$ column $40\%$ of the green rows contain $0$s, that is $60\%$ of the green rows in the $A$ column contain $1$s. Notice that the yellow rows belong to the $1$s in the $B$ column, and the green rows belong to the $0$s in the $B$ column. Also, we know that $50\%$ of the rows in the $A$ column are $1$s.

I just hope that it is clear that I described the question of the OP.

If we denote by $y$ the (unknown) number of the $1$s in the $B$ column, and by $g$ the number of the green columns then we can write that $$y+g=N,\tag 1$$ the number of the experiments. With this notation we can write that the number of the ones in the yellow rows of the $A$ column is $0.4y$ and that $0.6g$ is the number of the $1$s in the $A$ column belonging to the green rows. So $$0.4y+0.6g=0.5N\tag2$$ because $0.4y+0.6g$ gives the total number of the $1$s in the $A$ column.

From $(1)$ we have $g=N-y$. Substituting this to $(2$) we get that $$0.4y+0.6(N-y)=0.5N.$$ From this equation

$$y=0.5N.$$

That is, the number of the $1$s in the $B$ column is the half of the number of the experiments.

In terms of relative frequencies (being close to probabilities) we have

$$P(B)\approx \frac yN=0.5.$$

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In modern terms, the two equations, $(1)$ and $(2)$ can be written as follows:

$$P(B)+P(B')=1$$

and

$$P(A\mid B)P(B)+P(A\mid B')P(B')=P(A)$$

where $$P(A\mid B')=1-P(A'\mid B').\tag 3$$ We have two equations and the result for $P(B)$ is $0.5$.

Answer 2

Draw a Venn Diagram and label the regions: for example, let $$p(A\cap B)=x$$ and $$p(A'\cap B)=y$$

In which case, you also have $$p(A\cap B')=0.5-x$$ and $$p(A'\cap B')=0.5-y$$

You can then use the information about the conditional probabilities to obtain the following equations:

$$\frac{x}{x+y}=0.4$$ and $$\frac{0.5-y}{1-x-y}=0.4$$

From these you can easily obtain $x$ and $y$ and get the answer.

Answer 3

$$ P(A\cap B) = {2\over 5}P(B)$$ $$ 1- P(A\cup B)= P(A'\cap B') = {2\over 5}P(B')$$

if we sum these two we get $$1- P(A\cup B)+P(A\cap B) = {2\over 5}$$

and thus $$1-P(A)-P(B)= {2\over 5}$$

....