Question

If $A$ and $B$ are two events such that $P\left( {A \cup B} \right) = \dfrac{5}{6}$, $P\left( {A \cap B} \right) = \dfrac{1}{3}$, and $P\left( {B'} \right) = \dfrac{1}{3}$ then \[P\left( A \right)\] is:
A. $\dfrac{1}{4}$
B. $\dfrac{1}{3}$
C. $\dfrac{1}{2}$
D. $\dfrac{2}{3}$

Answer 1

Hint: Here, $P(x)$ denotes the probability of some event. Thus, $P\left( {A \cup B} \right)$ means the probability of $A \cup B$, that is either of the events A or B occur. Also, $P\left( {A \cap B} \right)$ means the probability of $A \cap B$, that is both A and B occur. So, we will make use of the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ to solve the problem and find the value of \[P\left( A \right)\].

Complete step by step solution:
So, we have, $P\left( {A \cup B} \right) = \dfrac{5}{6}$, $P\left( {A \cap B} \right) = \dfrac{1}{3}$, and $P\left( {B'} \right) = \dfrac{1}{3}$.
We have to find the value of \[P\left( A \right)\].
First, we will find the probability of event B by equating the sum of probabilities of all the possibilities to one.
Hence, $P\left( B \right) + P\left( {B'} \right) = 1$
$ \Rightarrow P\left( B \right) + \dfrac{1}{3} = 1$
$ \Rightarrow P\left( B \right) = 1 - \dfrac{1}{3}$
$ \Rightarrow P\left( B \right) = \dfrac{2}{3}$
Now, we will substitute the known values ion the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ to find the value of \[P\left( A \right)\].
So, we get,
$ \Rightarrow \dfrac{5}{6} = P(A) + \dfrac{2}{3} - \dfrac{1}{3}$
$ \Rightarrow \dfrac{5}{6} = P(A) + \dfrac{1}{3}$
$ \Rightarrow P(A) = \dfrac{5}{6} - \dfrac{1}{3} = \dfrac{3}{6}$
$ \Rightarrow P(A) = \dfrac{1}{2}$

Option ‘C’ is correct

Note: These problems are the combinations of sets and probability, so, the concepts of both of the topics are used in these. The formula $P(A + B) = P(A) + P(B) - P(AB)$ is widely used in such questions. This formula is a restructured version of the formula of sets, which is, $n(A \cup B) = n(A) + n(B) - n(A \cap B)$ where, $n(x)$denotes number of elements in set $x.$ This formula is modified into the formula of probability by dividing on both sides by $n(U)$, where, $U$ is the universal set.