Question

If A and B are two events such that $P\left( {AUB} \right) = \dfrac{3}{4}$, $P\left( {A \cap B} \right) = \dfrac{1}{4}$, $P\left( {A'} \right) = \dfrac{2}{3}$, then $P\left( {A' \cap B} \right)$ is equal to
A.$\dfrac{5}{{12}}$
B.$\dfrac{3}{8}$
C.$\dfrac{5}{8}$
D.$\dfrac{1}{2}$

Answer 1

Here the given question is based on the concept of probability. We have to find the value of $P\left( {A' \cap B} \right)$ using the given values. For this, we have to use the addition theorem of probability i.e., $P\left( {A' \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$ and given $P\left( {A'} \right)$ is the even that $A$ does not occur it can expressed as $P\left( A \right) = 1 - P\left( {A'} \right)$ and on further simplification we get the required solution.

Complete answer: The Addition theorem of probability define in two cases:
If two events A and B are mutually exclusive, then the probability of occurrence of either A or B is:
$P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right)$ -----(1)
If two events A and B are not mutually exclusive, then the probability of occurrence of either A or B is:
$P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$ ----(2)
Now, consider the given question:
If A and B are two events such that
$P\left( {AUB} \right) = \dfrac{3}{4}$
$P\left( {A \cap B} \right) = \dfrac{1}{4}$
$P\left( {A'} \right) = \dfrac{2}{3}$
Find $P\left( {A' \cap B} \right) = ?$
In $P\left( {A'} \right)$ the complement of $A$ or $A'$ means it consists of all the outcomes in which the event $A$ does not occur.
$P\left( A \right) + P\left( {A'} \right) = 1$
Or
$P\left( A \right) = 1 - P\left( {A'} \right)$
$\Rightarrow \,\,P\left( A \right) = 1 - \dfrac{2}{3}$
Take 3 as LCM in RHS, then
$\Rightarrow \,\,P\left( A \right) = \dfrac{{3 - 2}}{3}$
On simplification, we get
$\therefore \,\,P\left( A \right) = \dfrac{1}{3}$
Now consider the equation (2)
$\Rightarrow \,\,P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$
On rearranging, we can written as
$\Rightarrow \,\,\,P\left( B \right) = P\left( {A \cup B} \right) - P\left( A \right) + P\left( {A \cap B} \right)$
On substituting the values, we have
$\Rightarrow \,\,\,P\left( B \right) = \dfrac{3}{4} - \dfrac{1}{3} + \dfrac{1}{4}$
Take 12 as LCM in RHS, then
$\Rightarrow \,\,\,P\left( B \right) = \dfrac{{9 - 4 + 3}}{{12}}$
On simplification, we get
$\Rightarrow \,\,\,P\left( B \right) = \dfrac{8}{{12}}$
Divide both numerator and denominator by 4, then
$\therefore \,\,\,P\left( B \right) = \dfrac{2}{3}$
Now find $P\left( {A' \cap B} \right)$,
In $P\left( {A' \cap B} \right)$, $A' \cap B$ means in the event the only terms occur in B, then we have a formula
$P\left( {A' \cap B} \right) = P\left( B \right) - P\left( {A \cap B} \right)$
On substituting the values, we have
$\Rightarrow \,\,\,P\left( {A' \cap B} \right) = \dfrac{2}{3} - \dfrac{1}{4}$
Take 12 as LCM in RHS, then
$\Rightarrow \,\,\,P\left( {A' \cap B} \right) = \dfrac{{8 - 3}}{{12}}$
On simplification, we get
$\therefore \,\,\,P\left( {A' \cap B} \right) = \dfrac{5}{{12}}$
Hence, option (1) is the correct answer.

Note:
The probability is a number of possible values. Candidates must know the basic theorem that is addition and multiplication theorem. Remember the complement of an event is the event which is not occurring. If the probability that Event A will not occur is denoted by $P\left( {A'} \right)$ which is equal to $P\left( {A'} \right) = 1 - P\left( A \right)$.