Question

If $A$ and $B$ are two events, then

• A. $P(A\cap B)\le min\left\{P\right(A),P(B\left)\right\}$
• B. $P(A\cap B)\ge max\{0,1-P(A^{\prime })-P(B^{\prime }\left)\right\}$
• C. $P(A\cap B)\le P(A\cup B)$
• D. $P(A\cap B)=P\left(A\right)P\left(B\right)$ if $A$ and $B$ are independent

Answer 1

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

The correct options are
A $P(A\cap B)\le min\left\{P\right(A),P(B\left)\right\}$
B $P(A\cap B)\ge max\{0,1-P(A^{\prime })-P(B^{\prime }\left)\right\}$
C $P(A\cap B)\le P(A\cup B)$
D $P(A\cap B)=P\left(A\right)P\left(B\right)$ if $A$ and $B$ are independent

(a) It is obvious that the intersection between two sets is less than or equal to any of the sets. In fact, the intersection must be less than or equal to the minimum of the two sets.
(b) If sets $A$ and $B$ are disjoint, then intersection is $0$. Also, max of $0$ and $1-P\left(A^{\prime }\right)-P\left(B^{\prime }\right)$ is $0$ since the second term is negative. If, however, the sets intersect, $1-P\left(A^{\prime }\right)-P\left(B^{\prime }\right)=1-a-b-2r=p-r$. Hence, LHS $\ge$ RHS
(c) This is obvious
(d) This is obvious
Hemce, (a), (b), (c), (d) are correct.