Question

If A and B are two events, then which one of the following is not always true?

• A. $P(A\cap B)\ge P\left(A\right)+P\left(B\right)-1$
• B. $P(A\cap B)\le P\left(A\right)$
• C. $P(A^{\prime }\cap B^{\prime })\ge P\left(A^{\prime }\right)+P\left(B^{\prime }\right)-1$
• D. $P(A\cap B)=P\left(A\right).P\left(B\right)$

Answer 1

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

The correct options are
A $P(A\cap B)\ge P\left(A\right)+P\left(B\right)-1$
B $P(A\cap B)\le P\left(A\right)$
C $P(A^{\prime }\cap B^{\prime })\ge P\left(A^{\prime }\right)+P\left(B^{\prime }\right)-1$

A. $P(A\cap B)=P\left(A\right)+P\left(B\right)-P(A\cup B)$

$\Rightarrow$ if option A is always true, $P\left(A\right)+P\left(B\right)-P(A\cup B)\ge P\left(A\right)+P\left(B\right)-1$

$\Rightarrow P(A\cup B)\le 1$, which satisfies our assumption.

B. $P(A\cap B)\le P\left(A\right)$

This is always true, since the probability of occurences of variables in A has to be more than those in the intersection of A and B.

C. $P(A^{\prime }\cap B^{\prime })=P(A\cup B{)}^{\prime }=1-P(A\cup B)=1-P(A)-P(B)+P(A\cap B)$

If we assume $P(A^{\prime }\cap B^{\prime })\ge P\left(A^{\prime }\right)+P\left(B^{\prime }\right)-1$, we get $1-P\left(A\right)-P\left(B\right)+P(A\cap B)\ge 1-P\left(A\right)+1-P\left(B\right)-1$

$\Rightarrow P(A\cap B)\ge 0$, which satisfies our assumption. Hence this also is always true.

So, $P(A\cap B)=P\left(A\right)P\left(B\right)$

This is true only when A and B are independent of each other.

Hence, not always true.