Question

Let A and B be events for which \[P\left( A \right)=x,P\left( B \right)=y,P\left( A\cap B \right)=z\], then \[P\left( {{A}^{'}}\cap B \right)\] equals
A. $\left( 1-x \right)y$
B. $1-x+y$
C. $y-z$
D. $1-x+y-x$

Answer 1

Hint: We first explain the given probability forms of the \[P\left( A \right)=x,P\left( B \right)=y,P\left( A\cap B \right)=z\]. We try to express the \[P\left( {{A}^{'}}\cap B \right)\] with the help of set inclusion where the set \[\left( {{A}^{'}}\cap B \right)\] can be denoted as \[\left( {{A}^{'}}\cap B \right)=B\backslash A\]. We put the values to find the solution for \[P\left( {{A}^{'}}\cap B \right)\]. Then we use the concept of number of sets or the order of sets to establish the fact that solving the problem using probability concepts is similar to solving set theorems.

Complete step by step answer:
The given problem is the problem of probability in set inclusion.
There are two sets A and B.
\[P\left( A\cap B \right)=z\] denotes the probability of those two events happening at once.
We have to find the probability of \[P\left( {{A}^{'}}\cap B \right)\] which denotes the event A not happening and B happening at the same time.
We have to exclude the probability of \[P\left( A\cap B \right)\] from the probability of \[P\left( B \right)\].
In probability concept we have \[P\left( {{A}^{'}}\cap B \right)=P\left( B \right)-P\left( A\cap B \right)\].
So, \[P\left( {{A}^{'}}\cap B \right)=P\left( B \right)-P\left( A\cap B \right)=y-z\].
The correct option is C.
We know that fundamental theorem of probability gives us that \[P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}\] where \[n\left( S \right)\] denotes the universal set. To measure all the given probability, we would have used \[n\left( S \right)\].
This condition also applies for \[P\left( {{A}^{'}}\cap B \right)\].
Therefore, \[P\left( {{A}^{'}}\cap B \right)=\dfrac{n\left( {{A}^{'}}\cap B \right)}{n\left( S \right)}=\dfrac{n\left( B\backslash A \right)}{n\left( S \right)}\].
The set exclusion for order gives us that set B will be excluded only for the set \[P\left( A\cap B \right)\].
It gives \[P\left( {{A}^{'}}\cap B \right)=\dfrac{n\left( B\backslash A \right)}{n\left( S \right)}=\dfrac{n\left( B \right)-n\left( A\cap B \right)}{n\left( S \right)}=\dfrac{n\left( B \right)}{n\left( S \right)}-\dfrac{n\left( A\cap B \right)}{n\left( S \right)}=P\left( B \right)-P\left( A\cap B \right)\].

Note:
We need to remember that the universal set is similar for all the given probabilities like \[P\left( A \right),P\left( B \right),P\left( A\cap B \right),P\left( {{A}^{'}}\cap B \right)\]. That’s why we didn’t use the concept of number of points in the set and instead we directly used the probability form to find the solution.