Question

Let $P\left(E\right)$ denote the probability of occurrence of event $E.$ If $A$ and $B$ are two events associated with an experiment such that $\left|\begin{array}{ccc}P\left(A\right)-1& P\left(B\right)& -P(A\cap B)\\ P\left(B\right)& -P(A\cap B)& P\left(A\right)-1\\ -P(A\cap B)& P\left(A\right)-1& P\left(B\right)\end{array}\right|=0,$ then

• A. $P(A\cup B)=1$
• B. $P\left(A|B\right)=\frac{P\left(A\right)+P\left(B\right)-1}{P\left(B\right)}$
• C. If $A\subset B,$ then $P\left(B\right)=P(A\cup B)$
• D. If $A\subset B,$ then $P\left(A\right)=1$

Answer 1

Answer: (A), (B), (C)

If $A\subset B,$ then $P\left(B\right)=P(A\cup B)$

Let $a=P\left(A\right)-1,$ $b=P\left(B\right)$ and $c=-P(A\cap B)$

$\left|\begin{array}{ccc}P\left(A\right)-1& P\left(B\right)& -P(A\cap B)\\ P\left(B\right)& -P(A\cap B)& P\left(A\right)-1\\ -P(A\cap B)& P\left(A\right)-1& P\left(B\right)\end{array}\right|=0\Rightarrow \left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=0\Rightarrow 3abc-a^3-b^3-c^3=0$
$\Rightarrow a+b+c=0$ or $a=b=c$ (not possible)

From $a+b+c=0,$
$P\left(A\right)-1+P\left(B\right)-P(A\cap B)=0$
$\Rightarrow P(A\cup B)=1$

Now,
$P\left(A|B\right)=\frac{P(A\cap B)}{P\left(B\right)}$
$\Rightarrow \frac{P\left(A\right)+P\left(B\right)-P(A\cup B)}{P\left(B\right)}$
$=\frac{P\left(A\right)+P\left(B\right)-1}{P\left(B\right)}$

Again, if $A\subset B,$
then $A\cup B=B$
$\Rightarrow P(A\cup B)=P\left(B\right)=1$

If $A\subset B,$ then $P\left(A\right)<P\left(B\right)$
So, $P\left(A\right)$ can't be $1$