Question

If ${\text{P(A/B) = P(B/A)}}$, A and B are two non- mutually exclusive events then-
A.A and B are necessarily same events
B.P(A)= P(B)
C.${\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) = {\text{P(A)P(B)}}$
D.All of the above

Answer 1

Hint: Here, we will use multiplication theorem of conditional probability according to which $ \Rightarrow {\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) = {\text{P}}\left( {\text{A}} \right){\text{P}}\left( {\dfrac{{\text{B}}}{{\text{A}}}} \right)$ only if P (A) ≠$0$ And ${\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) = {\text{P}}\left( {\text{B}} \right){\text{P}}\left( {\dfrac{{\text{A}}}{{\text{B}}}} \right)$only if P (B) ≠$0$
Now use these two theorems and put the values in the given equation. Solve the obtained eq. to get the answer. Since A and B are non-mutually exclusive events so they will have at least one thing in common between them and the events cannot prevent the occurrence of each other.

Complete step-by-step answer:
It is given that ${\text{P}}\left( {\dfrac{{\text{A}}}{{\text{B}}}} \right){\text{ = P}}\left( {\dfrac{{\text{B}}}{{\text{A}}}} \right)$-- (i)
We know from multiplication theorem that If A and B are two events associated with a random experiment then,
$ \Rightarrow {\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) = {\text{P}}\left( {\text{A}} \right){\text{P}}\left( {\dfrac{{\text{B}}}{{\text{A}}}} \right)$ only if P (A) ≠$0$ -- (ii)
And ${\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) = {\text{P}}\left( {\text{B}} \right){\text{P}}\left( {\dfrac{{\text{A}}}{{\text{B}}}} \right)$only if P (B) ≠$0$ -- (iii)
From eq. (i) and (ii), we can write-
$ \Rightarrow {\text{P}}\left( {\dfrac{{\text{A}}}{{\text{B}}}} \right) = \dfrac{{{\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right)}}{{{\text{P}}\left( {\text{B}} \right)}}$and $ \Rightarrow {\text{P}}\left( {\dfrac{{\text{B}}}{{\text{A}}}} \right) = \dfrac{{{\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right)}}{{{\text{P}}\left( {\text{A}} \right)}}$-- (iv)
Then from eq. (i) and eq. (iv), we can write-
$ \Rightarrow \dfrac{{{\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right)}}{{{\text{P}}\left( {\text{B}} \right)}} = \dfrac{{{\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right)}}{{{\text{P}}\left( {\text{A}} \right)}}$
On cross multiplication, we get-
$ \Rightarrow {\text{P}}\left( {\text{A}} \right){\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) = {\text{P}}\left( {\text{B}} \right){\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right)$
On canceling the same terms on both sides, we get-
$ \Rightarrow $P (A)=P (B)
So option B is correct.
Here the probabilities are the same. The events A and B may differ but they are non-mutually exclusive events, they will have at least one common outcome between them So events A and B are not necessarily the same events.
Option A is incorrect.
Now, we know that${\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) = {\text{P}}\left( {\text{A}} \right){\text{P}}\left( {\text{B}} \right)$ only if A and B are independent but here A and B have something common in them so they are not independent so
$ \Rightarrow {\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) \ne {\text{P(A)P(B)}}$
So option C is incorrect.
Since all options are not correct, so option D is also incorrect.
The correct answer is option ‘B’.

Note: For non-mutually exclusive event the union of events A and B is given as-$ \Rightarrow {\text{P}}\left( {{\text{A}} \cup {\text{B}}} \right) = {\text{P}}\left( {\text{A}} \right){\text{ + P}}\left( {\text{B}} \right){\text{ - P}}\left( {{\text{A}} \cap {\text{B}}} \right)$
Mutually exclusive events are such events that cannot occur simultaneously and they have nothing common between them so the union for two mutually exclusive events is calculated as-$ \Rightarrow {\text{P}}\left( {{\text{A}} \cup {\text{B}}} \right) = {\text{P}}\left( {\text{A}} \right){\text{ + P}}\left( {\text{B}} \right)$