Question

If $P(A)=\dfrac{1}{2}$ , \[P(B)=\dfrac{1}{3}\] and\[P(A\cap B)=\dfrac{7}{12}\], then the value of \[P({{A}^{\prime }}\cap {{B}^{'}})\] is
A. \[\dfrac{7}{12}\]
B. \[\dfrac{3}{4}\]
C. \[\dfrac{1}{4}\]
D. \[\dfrac{1}{6}\]

Answer 1

Hint: In this question, we are to find the probability of the intersection of complimented events. By using set theory (complement of sets) and addition theorem on probability, the required probability is calculated.



Formula Used:The probability is calculated by,
\[P(E)=\dfrac{n(E)}{n(S)}\]
\[n(E)\] is the number of favourable outcomes and \[n(S)\] is the total number of outcomes.
If there are two events in a sample space, then the addition theorem on probability is given by
\[P(A\cup B)=P(A)+P(B)-P(A\cap B)\]
According to set theory, the complement of the union/intersection of two sets is equal to the intersection/union of complemented sets respectively. I.e.,
$\begin{align}
  & (\overline{A\cup B})={{A}^{'}}\cap {{B}^{'}} \\
 & (\overline{A\cap B})={{A}^{'}}\cup {{B}^{'}} \\
\end{align}$



Complete step by step solution:Consider two events $A$ and \[B\].
Given that,
$P(A)=\dfrac{1}{2}$
\[P(B)=\dfrac{1}{3}\]
\[P(A\cap B)=\dfrac{7}{12}\]
According to set theory,
According to set theory, the complement of the union of two sets is equal to the intersection of complemented sets. I.e.,
$(\overline{A\cup B})={{A}^{'}}\cap {{B}^{'}}$
So, the required probability is,
\[P({{A}^{\prime }}\cap {{B}^{'}})=P(\overline{A\cup B})\]
But we know
\[P(\overline{A\cup B})=1-P(A\cup B)\]
Then,
\[\begin{align}
  & P({{A}^{\prime }}\cap {{B}^{'}})=P(\overline{A\cup B}) \\
 & \text{ }=1-P(A\cup B) \\
\end{align}\]
According to the addition theorem on probability,
\[P(A\cup B)=P(A)+P(B)-P(A\cap B)\]
On substituting,
\[\begin{align}
  & P({{A}^{\prime }}\cap {{B}^{'}})=P(\overline{A\cup B}) \\
 & \text{ }=1-P(A\cup B) \\
 & \text{ }=1-\left[ P(A)+P(B)-P(A\cap B) \right] \\
 & \text{ }=1-\left[ \dfrac{1}{2}+\dfrac{1}{3}-\dfrac{7}{12} \right] \\
\end{align}\]
On simplifying, we get
\[\begin{align}
  & P({{A}^{\prime }}\cap {{B}^{'}})=1-\left[ \dfrac{6+4-7}{12} \right] \\
 & \text{ }=1-\dfrac{3}{12} \\
 & \text{ }=1-\dfrac{1}{4} \\
 & \text{ }=\dfrac{3}{4} \\
\end{align}\]



Option ‘B’ is correct

Note: Here we may go wrong with the calculation of $P({{A}^{'}}\cap {{B}^{'}})$. Where the required probability is the intersection of complements. So, we get a complement of the union of sets in the result. Here the set theory helps to find such a type of complemented set. By substituting the appropriate values, the required probability is calculated.