Question

If $A$ and $B$ are two events such that $P(A \cup B) = \dfrac{5}{6}$, $P(A \cap B) = \dfrac{1}{3}$, $P(B') = \dfrac{1}{3}$, then $P(A) = $
(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, $P(A \cup B)$ means the probability of $A \cup B$. Similarly, $P(A \cap B)$ means the probability of $A \cap B$, and $P(B')$ is the probability of $B'$. So, we are to find the probability of$A$, i.e., $P(A)$. To find,$P(A)$, we will use the formula of probability,
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
And, to find, $P(B)$, we will use the formula,
$P(B) = 1 - P(B')$

Complete answer:
In the given question, we are provided with the values $P(A \cup B) = \dfrac{5}{6}$, $P(A \cap B) = \dfrac{1}{3}$ and $P(B') = \dfrac{1}{3}$.
We know that the sum of probabilities of all the possibilities of an event is one. So, to find the probability of $B$, we will use the formula,
\[P(B) = 1 - P(B') - - - - (1)\]
Substituting, the value of $P(B')$ in $(1)$, we get,
$ \Rightarrow P(B) = 1 - \dfrac{1}{3}$
Taking the LCM of the denominators, we get,
$ \Rightarrow P(B) = \dfrac{{3 - 1}}{3}$
$ \Rightarrow P(B) = \dfrac{2}{3}$
Now, to find probability of $A$, that is,$P(A)$, we will use the formula,
$P(A \cup B) = P(A) + P(B) - P(A \cap B) - - - - (2)$
Now, substituting the values of $P(A \cup B)$, $P(A \cap B)$ and $P(B)$ in $(2)$, we get,
$ \Rightarrow \dfrac{5}{6} = P(A) + \dfrac{2}{3} - \dfrac{1}{3}$
Now, adding $\dfrac{1}{3}$ in both sides of the equation, we get,
$ \Rightarrow \dfrac{5}{6} + \dfrac{1}{3} = P(A) + \dfrac{2}{3}$
Cancelling the like terms with opposite signs, we get,
$ \Rightarrow \dfrac{{5 + 2}}{6} = P(A) + \dfrac{2}{3}$
$ \Rightarrow \dfrac{7}{6} = P(A) + \dfrac{2}{3}$
Now, subtracting $\dfrac{2}{3}$from both sides of the equation, we get,
$ \Rightarrow \dfrac{7}{6} - \dfrac{2}{3} = P(A)$
Taking LCM of both the denominators, we get,
$ \Rightarrow \dfrac{{7 - 4}}{6} = P(A)$
Simplifying the expression, we get,
$ \Rightarrow \dfrac{3}{6} = P(A)$
Cancelling the common factors in numerator and denominator, we get,
$ \Rightarrow P(A) = \dfrac{1}{2}$
Therefore, the probability of $A$ is $P(A) = \dfrac{1}{2}$, that is, option C.
Hence, option (B) is the correct answer.

Note:
 These problems are the combinations of sets and probability, so, the concepts of both of the topics are used in these. Here the formula, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ is used. 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. On dividing each term by $n(U)$, it takes the form, $\dfrac{{n(A \cup B)}}{{n(U)}} = \dfrac{{n(A)}}{{n(U)}} + \dfrac{{n(B)}}{{n(U)}} - \dfrac{{n(A \cap B)}}{{n(U)}}$. Now, we know, ${\text{Probability = }}\dfrac{{{\text{No}}{\text{. of favourable outcomes}}}}{{{\text{Total no}}{\text{. of outcomes}}}}$. The same concept is used above, and the following formula is derived,
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.