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.
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.
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 is consistent with the conditional probabilities, and any value satisfying $0 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.
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 $P(\overline{A})=1-P(A).$)