Question

Suppose that $P\left(A\right)=3/5$ and $P\left(B\right)=2/3$. Then

• A. $P(A\cup B)\le 2/3$
• B. $4/15\le P(A\cap B)\le 3/5$
• C. $2/5\le P\left(A|B\right)\le 9/10$
• D. $P(A\cap B^{\prime })\le 1/3$

Answer 1

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

The correct options are
B $4/15\le P(A\cap B)\le 3/5$
C $2/5\le P\left(A|B\right)\le 9/10$
D $P(A\cap B^{\prime })\le 1/3$

Maximum of $P(A\cup B)$ can be 1.
Maximum of $P(A\cap B)$ is minimum of $A$ and $B$, hence, $3/5$.
Minimum of $P(A\cap B)$ is when the union of $A$ and $B$ is 1. Thus, $P(A\cap B)$ = $2/3+3/5-1=4/15$
Thus, $P\left(A|B\right)=P(A\cap B)/P\left(B\right)$ will range from $\frac{\frac{3}{5}}{\frac{2}{3}}=\frac{9}{10}$ and $\frac{\frac{4}{15}}{\frac{2}{3}}=\frac{2}{5}$
Maximum of $P(A\cap B^{\prime })$ is maximum when there is minimum intersection, hence $P(A\cap B^{\prime })=P\left(A\right)-P(A\cap B)=\frac{3}{5}-\frac{4}{15}=\frac{1}{3}$
Hence, (B), (C), (D) are correct.