Question

If two events $A$ and \[B\] are such that \[P(A+B)=\dfrac{5}{6}\], \[P(A\cap B)=\dfrac{1}{3}\] and\[P(\overline{A})=\dfrac{1}{2}\], then the events $A$ and \[B\] are
A. Independent
B. Mutually exclusive
C. Mutually exclusive and independent
D. None of these

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
\[P(A+B)=P(A)+P(B)-P(A\cap B)\]
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 independent events if $P(A\cap B)=P(A)\cdot P(B)$ and are said to be mutually exclusive if $P(A\cap B)=\Phi $.

Complete step by step solution: Consider two events $A$ and \[B\].
It is given that,
\[P(A+B)=\dfrac{5}{6}\]
\[P(A\cap B)=\dfrac{1}{3}\]
\[P(\overline{A})=\dfrac{1}{2}\]
\[\begin{align}
  & P(A)=1-P(\overline{A}) \\
 & \text{ }=1-\dfrac{1}{2} \\
 & \text{ }=\dfrac{1}{2} \\
\end{align}\]
The addition theorem on probability is given by

Then, by substituting in the formula, we get
\[\begin{align}
  & P(A+B)=P(A)+P(B)-P(A\cap B) \\
 & \text{ }\dfrac{5}{6}=\dfrac{1}{2}+P(B)-\dfrac{1}{3} \\
 & \text{ }\Rightarrow P(B)=\dfrac{5}{6}-\dfrac{1}{6}=\dfrac{2}{3} \\
\end{align}\]
The given two events are independent if and only if $P(A\cap B)=P(A)\cdot P(B)$
Here we have $P(A)=\dfrac{1}{2};P(B)=\dfrac{2}{3}$
Then,
\[\begin{align}
  & P(A)\cdot P(B)=\dfrac{1}{2}\times \dfrac{2}{3} \\
 & \text{ }=\dfrac{1}{3} \\
 & \text{ }=P(A\cap B) \\
\end{align}\]
Thus, they are independent events.

Option ‘A’ is correct

Note: Here we may go wrong with the value of $P(A\cap B)$. For independent events $P(A\cap B)\ne \Phi $. This means, there is no intersection between the two events.