Question

If \[A\] and \[B\] are any two events, then \[P(\overline{A}\cap B)=\]
A. \[P(\overline{A})\cdot P(\overline{B})\]
B. \[1-P(A)-P(B)\]
C. \[P(A)+P(B)-P(A\cap B)\]
D. \[P(B)-P(A\cap B)\]

Answer 1

Hint: In the above question, we are to find the value of \[P(\overline{A}\cap B)\]. 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,
\[P(E)=\dfrac{n(E)}{n(S)}\]
\[n(E)\] - favorable 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,
The union of sets is described as:
\[A\cup B=\{x/x\in A\text{ or }x\in B\}\]
The intersection of sets is described as:
\[A\cap B=\{x/x\in A\text{ and }x\in B\}\]
The symmetric difference is described as:
\[\begin{align}
  & A-B=\{x/x\in A\text{ and }x\notin B\} \\
 & B-A=\{x/x\in B\text{ and }x\notin A\} \\
\end{align}\]



Complete step by step solution:Given that, \[A\] and \[B\] are any two events.
According to set theory, the addition theorem on probability is,
\[P(A\cup B)=P(A)+P(B)-P(A\cap B)\]
Then,
\[P(A\cup B)-P(A)=P(B)-P(A\cap B)\]
Here we can write,
\[P(A\cup B)-P(A)=P(\overline{A}\cap B)\]
Therefore,
\[P(\overline{A}\cap B)=P(B)-P(A\cap B)\]



Option ‘D’ is correct

Note: Here we may go wrong with the set theory formula. If the Venn diagram is used, then the formulae will be established easily. Then, by applying the probability to the sets, the required addition theorem on probability is obtained. 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.