Question

The probabilities of three mutually exclusive events are $\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.,\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ and $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.$. The statements is

• A. True
• B. Wrong
• C. Could be either
• D. Do not know

Answer 1

The correct option is B

Wrong

Explanation for the correct option:

Given that the probabilities of $3$ mutually exclusive events are $\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.,\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ and $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.$

Let us assume that $A,B$ and $C$ are the three events.

Then, $P(A+B+C)=P\left(A\right)+P\left(B\right)+P\left(C\right)$

$=\left(\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)+\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right)+\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.\right)$

$=\raisebox{1ex}{$13$}\!\left/ \!\raisebox{-1ex}{$12$}\right.$ which is greater than $1$

Hence, the given statement is wrong.

Hence, the correct answer is option (B), Wrong.