Question

IIf A and B are two events such that\[P\left( {A \cup B} \right){\text{ }} = {\text{ }}P\left( {A{\text{ }} \cap {\text{ }}B} \right)\], then the true relation is
\[\left( 1 \right)\] \[P\left( A \right) + P\left( B \right) = 0\]
\[\left( 2 \right)\] \[P\left( A \right) + P\left( B \right) = P\left( A \right)P\left( {B|A} \right)\]
\[\left( 3 \right)\] \[P\left( A \right) + P\left( B \right) = 2P\left( A \right)P\left( {B|A} \right)\]
$(4)$none of these

Answer 1

Hint: We have to find the relation which is similar to the given relation i.e. \[P\left( {A \cup B} \right) = P\left( {A \cap B} \right)\] . We solve this question using the property of conditional probability of an event \[E\] , given that the occurrence of the event F is given by\[P\left( {E|F} \right) = \dfrac{{P(E \cap F)}}{{P(F)}}{\text{ }}\],$P(F) \ne 0$. Also using the formulas of probability stating the relation between union and intersection of two events we find the relation equivalent to the given relation .

Complete step-by-step answer:
The intersection of two terms or events are said to the common elements shared by the two events . The symbol of intersection is also stated as “ and “ . whereas the union of the two terms or events is said to be the sum total of the two events and subtracting the common portion or the intersection of the two events or terms . The symbol of union is also stated as “ or “ .

Given : \[P\left( {A \cup B} \right) = P\left( {A \cap B} \right)\]
We know ,
\[P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)-----\left( 1 \right)\]
Substituting the given condition in equation \[\left( 1 \right)\] , we get
\[P\left( {A \cap B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\]
On further simplifying , we get
\[2{\text{ }}P\left( {A \cap B} \right) = P\left( A \right) + P\left( B \right)------\left( 2 \right)\]
Also , we know
\[P\left( {B|A} \right) = \dfrac{{P(A \cap B)}}{{P(A)}}------\left( 3 \right)\]
From \[\left( 3 \right)\] we get
\[P\left( {A \cap B} \right) = P\left( {B|A} \right) \times P\left( A \right)-----\left( 4 \right)\]
Putting the value of \[\left( 4 \right)\] in equation \[\left( 2 \right)\] , we get
\[2P\left( {B|A} \right) \times P\left( A \right) = P\left( A \right) + P\left( B \right)\]
Thus , the correct option is \[\left( 3 \right)\] .
So, the correct answer is “Option 3”.

Note: We find the relation by using the conditional probability formula . If \[A\] and \[B\] are mutually exclusive events , then\[P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right)\]. The basic property of probability is that the probability of an event can never be greater than \[1\] .
If two events A and B are independent , then
\[P\left( {E \cap F} \right) = P\left( E \right) \times P\left( F \right)\]
\[P\left( {E|F} \right) = P\left( E \right){\text{ }}\], $P(F) \ne 0$
\[P\left( {F|E} \right) = P\left( F \right){\text{ }}\],$P(F) \ne 0$