Question

Suppose that $A,B,C$ are events such that $P\left(A\right)=P\left(B\right)=P\left(C\right)=\frac{1}{4}$, $P\left(AB\right)=0$, $P\left(AC\right)=\frac{1}{8}$, then $P(A+B)=$

• A. $0.125$
• B. $0.25$
• C. $0.375$
• D. $0.5$

Answer 1

The correct option is D

$0.5$

Explanation for the correct option :

Calculating $P(A+B)$ :

Given that $P\left(A\right)=P\left(B\right)=P\left(C\right)=\frac{1}{4}$ and $P\left(AB\right)=0$, $P\left(AC\right)=\frac{1}{8}$.

Formula to be used : We know that for any two events $A$ and $B$, $P(A+B)=P\left(A\right)+P\left(B\right)-P\left(AB\right)$.

So, using the given values, we get

$$\begin{gathered} P(A+B)=P\left(A\right)+P\left(B\right)-P\left(AB\right) \\ =\frac{1}{4}+\frac{1}{4}-0 \\ =\frac{1}{2} \\ =0.5 \end{gathered}$$

Hence, option (D) is the correct answer.