Question

The probabilities of three mutually exclusive events are $\dfrac{2}{3}$, $\dfrac{1}{4}$ and $\dfrac{1}{6}$. The given statement is
A. True
B. Wrong
C. Could be either
D. Do not know

Answer 1

Hint: In this question, we have to find the relationship between the events. Here we have a few values, we can use to find the relationship between the given events. For this question, the addition theorem on probability is used. All the given values are substituted in the addition theorem of probability to find the required probability

Formula Used:A probability is the ratio of favorable outcomes of an event to the total number of outcomes. So, the probability lies between 0 and 1.
The probability is calculated by,

 is the number of favorable outcomes and is the total number of outcomes.
If there are two events in a sample space, then the addition theorem on probability is given by

In independent events, the occurrence of one event is not affected by the occurrence of another event.
Two events $A$ and \[B\] are said to be mutually exclusive if $P(A\cap B)=\Phi $.
If two events $A$ and \[B\] are mutually exclusive, then $P(A\cup B)=P(A)+P(B)$

Complete step by step solution: Consider three events $A$, \[B\] and $C$.
It is given that,
\[P(A)=\dfrac{2}{3}\]
\[P(B)=\dfrac{1}{4}\]
\[P(C)=\dfrac{1}{6}\]
The given three events are mutually exclusive if and only if $P(A\cup B\cup C)=P(A)+P(B)+P(C)$
$\begin{align}
  & P(A\cup B\cup C)=\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{1}{6} \\
 & \text{ }=\dfrac{8+3+2}{12} \\
 & \text{ }=\dfrac{13}{12} \\
\end{align}$
Since the obtained probability is greater than 1. So, the given three events are not exclusive events.

Option ‘B’ is correct

Note: Here we may go wrong with the value of $P(A\cup B)$. For mutually exclusive events $P(A\cap B)=\Phi $. Then, there is no intersection between the two events and $P(A\cup B)=P(A)+P(B)$.