Question

If \[P(A)=0.25\], \[P(B)=0.50\] and \[P(A\cap B)=0.14\] , then \[P(A\cap \overline{B})\] is equal to
A. \[0.61\]
B. \[0.39\]
C. \[0.48\]
D. None of these

Answer 1

Hint: In this question, we are to find the probability of an event. 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,
\[P(E)=\dfrac{n(E)}{n(S)}\]
\[n(E)\] - favourable outcomes and \[n(S)\] - sample.
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,
\[\begin{align}
  & A\cup B=\{x/x\in A\text{ or }x\in B\} \\
 & A\cap B=\{x/x\in A\text{ and }x\in B\} \\
 & A-B=\{x/x\in A\text{ and }x\notin B\} \\
 & B-A=\{x/x\in B\text{ and }x\notin A\} \\
\end{align}\]



Complete step by step solution:Given that,
\[P(A)=0.25\]
\[P(B)=0.50\]
\[P(A\cap B)=0.14\]
According to set theory,
The probability of finding an element in a given set is equal to the sum of probabilities of finding elements in an exclusive part of that set and those parts that contain elements that are common to other sets also.
By the addition theorem on probability,
\[\begin{align}
  & P(A\cup B)=P(A)+P(B)-P(A\cap B) \\
 & \Rightarrow P(A\cup B)-P(B)=P(A)-P(A\cap B) \\
\end{align}\]
We can write
\[P(A\cup B)-P(B)=P(A\cap \overline{B})\]
Then, the required probability is
$\begin{align}
  & P(A\cap \overline{B})=P(A)-P(A\cap B) \\
 & \text{ }=0.25-0.14 \\
 & \text{ }=0.11 \\
\end{align}$



Option ‘D’ is correct

Note: In this question, the addition theorem on probability is applied for finding the required probability. By substituting the appropriate values, the required probability is calculated. Here we may go wrong with the complimented probability.