Question

Given that $A,B$ and $C$ are events such that $P\left(A\right)=P\left(B\right)=P\left(C\right)=1/5,P(A\cap B)=P(B\cap C)=0$ and $P(A\cap C)=1/10$. The probability that at least one of the events $A,B$ or $C$ occurs is

• A. $1/2$
• B. $2/3$
• C. $2/5$
• D. $3/5$

Answer 1

The correct option is A $1/2$

We have $A\cap B\cap C\underset{–}{\cap }A\cap B$
$\Rightarrow P(A\cap B\cap C)\le P(A\cap B)=0$
Since, $P(A\cap B\cap C)\ge 0$, we get $P(A\cap B\cap C)=0$.
Now, $P$ (at least one of $A,B,C$)
$=P(A\cap B\cap C)=P\left(A\right)+P\left(B\right)+P\left(C\right)-P(B\cap C)-P(C\cap A)-P(A\cap B)+P(A\cap B\cap C)$
$\frac{1}{5}+\frac{1}{5}+\frac{1}{5}-0-\frac{1}{10}-0+0=\frac{1}{2}$.