# Probability rules with two conditionals only

I have an homework with these two conditions :

P(B|A) = 0.87
P(¬B|¬A) = 0.98

And I have to find P(B). ....I have no idea how to procede from there.

Both events A, and B are dependent and not mutually exclusive.

Any help appreciated.

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You can't get a numerical value; the best you can do is get $P(B)$ in terms of $P(A)$.

If $a=P(A)$, then $P(A\land B)=0.87a$, so $P(A\land\lnot B)=a-0.87a=0.13a$. We also have $P(\lnot A)=1-a$, so $P(\lnot A\land\lnot B)=0.98(1-a)$, and $P(\lnot A\land B)=0.02(1-a)$.

Now $P(B)=P(B\land A)+P(B\land\lnot A)=0.87a+0.02(1-a)=0.02+0.85a$. Any value of $a$ satisfying $0<a\le 1$ is consistent with the conditional probabilities, and any value satisfying $0<a<1$ also makes $A$ and $B$ dependent. For example, if $a=P(A)=0.20$, then $P(B)=0.19$, while if $P(A)=0.40$, then $P(B)=0.36$. In the first case the probabilities of the four possible combinations are:

$$\begin{array}{r|c} &A&\lnot A&\text{Total}\\ \hline B&0.174&0.016&0.19\\ \hline \lnot B&0.026&0.784&0.81\\ \hline \text{Total}&0.20&0.80&1.00 \end{array}$$

In the second:

$$\begin{array}{r|c} &A&\lnot A&\text{Total}\\ \hline B&0.348&0.012&0.36\\ \hline \lnot B&0.052&0.588&0.64\\ \hline \text{Total}&0.40&0.60&1.00 \end{array}$$

As you can see, both are perfectly consistent, as is any other such combination.

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You have my respect. Haha! I'll digest that this morning. Thank you for the thorough explaination. I'll surely note this and keep it safe! –  Yanick Rochon May 11 '12 at 9:44
I'm afraid you cannot find $P(B)$ without knowing $P(A)$. If you do know this, you can use the expression $P(B)=P(B|A)P(A)+P(B|\overline{A})P(\overline{A})$ and the fact that, since $P(\cdot|\overline{A})$ is a probability, $P(B|\overline{A})=1-P(\overline{B}|\overline{A}).$ (Edit: And of course $P(\overline{A})=1-P(A).$)