Question

For any 2 events A and B in a sample space which one of the following is incorrect?

• A. $P\left(A/B\right)\ge \frac{P\left(A\right)+P\left(B\right)-1}{P\left(B\right)},P\left(B\right)\ne 0$, is always true
• B. $P(A\cap \overline{B})=P\left(A\right)-P(A\cap B)$ doen't hold
• C. $P(A\cup B)=1-P\left(\overline{A}\right)P\left(\overline{B}\right)$ if A & B are independent
• D. $P(A\cup B)=1-P\left(\overline{A}\right)P\left(\overline{B}\right)$, if A & B are disjoint.

Answer 1

The correct option is A $P\left(A/B\right)\ge \frac{P\left(A\right)+P\left(B\right)-1}{P\left(B\right)},P\left(B\right)\ne 0$, is always true

Consider $2$ events $A$ $\xi$ $B$

$P\left(\frac{A}{B}\right)\ge \frac{P\left(A\right)+P\left(B\right)-1}{P\left(B\right)},P\left(B\right)\ne 0,$ is always true.

This condition is not correct.

$P\left(\frac{A}{B}\right)=\frac{NumberofelementaryeventsfavorableA\cap B}{NumberofelementaryeventsfavorabletoB}$

$P\left(\frac{A}{B}\right)=\frac{n(A\cap B)}{n\left(B\right)}=\frac{\frac{n(A\cap B)}{n\left(s\right)}}{\frac{n\left(B\right)}{n\left(s\right)}}$

$\therefore P\left(\frac{A}{B}\right)=\frac{(A\cap B)}{\left(B\right)}$

Similarly $P\left(\frac{B}{A}\right)=\frac{(A\cap B)}{\left(A\right)}$

Hence, the answer is $P\left(\frac{A}{B}\right)\ge \frac{P\left(A\right)+P\left(B\right)-1}{P\left(B\right)},P\left(B\right)\ne 0,$ is always true.