Question

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

• A. $P(A\cap B)\ge P\left(A\right)+P\left(B\right)$
• B. $P(A\cup B)\le P\left(A\right)+P\left(B\right)$
• C. $P(A\cap B)=P\left(A\right)+P\left(B\right)$
• D. None of these

Answer 1

The correct option is B

$P(A\cup B)\le P\left(A\right)+P\left(B\right)$

Explanation for the correct option:

Finding the probability:

Consider two sets $AandB$

Then by the addition theorem of probability,

$\mathit{P}\mathbf{(}\mathit{A}\mathbf{\cup }\mathit{B}\mathbf{)}\mathbf{=}\mathit{P}\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{+}\mathit{P}\mathbf{\left(}\mathit{B}\mathbf{\right)}\mathbf{-}\mathit{P}\mathbf{(}\mathit{A}\mathbf{\cap }\mathit{B}\mathbf{)}$

$\begin{array}{rcl}\because P(A\cap B)& \ge & 0\\ \therefore P\left(A\right)+P\left(B\right)& \ge & P(A\cup B)\\ & & \end{array}$

Hence, option$\mathbf{\left(}\mathit{B}\mathbf{\right)}$ is the correct.