Question

If $\text{A}$ and $\text{B}$ are two events such that $\text{P(A) =}\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$, and $\text{P(B) =}\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$, then

• A. $\text{P(AUB) ≥ 2/3}$
• B. $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.\text{≤P(A∩B)≤ 1/2}$
• C. $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.\text{≤ P(A'∩B) ≤1/2}$
• D. All of these

Answer 1

The correct option is D

All of these

Step 1. Find the value of $\mathbf{\text{P(AUB)}}$

Given data,

$\text{P}\left(\text{A}\right)$ $=$ $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

$\text{P}\left(\text{B}\right)$ $=$ $\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$

We know that, $\text{P(AUB)}$ $\ge$ max $\left\{\text{P(A),P(B)}\right\}$ $=$ $\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$

Step 2. Find the value of $\mathbf{\text{P(A∩B)}}$

We know that, $\text{P(A∩B)}$ $\le$ min $\left\{\text{P(A),P(B)}\right\}$ $=$$\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

As $\mathbf{\text{P(A∩B)}}$ $\mathbf{=}$ $\mathbf{\text{P}}\mathbf{\left(}\mathbf{\text{A}}\mathbf{\right)}\mathbf{\text{+ P(B) - P(A∪B)}}$

$\ge$ $\text{P}\left(\text{A}\right)$$+$$\text{P}\left(\text{B}\right)$$-$$1$

$\ge$ $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ $+$ $\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$ $-$$1$

$\ge$ $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.$

Therefore, $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.$$\le$$\text{P(A∩B)}$$\le$$\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

Step 3. Find the value of $\mathbf{\text{P(A'∩B)}}$

We know that, $\mathbf{\text{P(A'∩B)}}$$\mathbf{=}$$\mathbf{\text{P}}\left(\text{B}\right)$$\mathbf{-}$$\mathbf{\text{P(A∩B)}}$

So, $\left(\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)$$-$$\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$ $\le$ $\text{P(A'∩B)}$$\le$ $\left(\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)$$-$$\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.\right)$

$\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.\right)$ $\le$$\text{P(A'∩B)}$$\le$$\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$

Hence, all the given statements are true.

Therefore, option $"\text{D"}$ is correct.