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Question: If \(A\) and \(B\) are two events such that \(A \subset B\) and \(P\left( B \right) \ne 0\), then wh...

If AA and BB are two events such that A⊂BA \subset B and P(B)≠0P\left( B \right) \ne 0, then which of the following is correct?
A) (a) P(A∣B)=P(B)P(A)\left( a \right){\text{ P}}\left( {A\left| B \right.} \right) = \dfrac{{P\left( B \right)}}{{P\left( A \right)}}
B) (b) P(A∣B)<P(A)\left( b \right){\text{ P}}\left( {A\left| B \right.} \right) < P\left( A \right)
C) (c) P(A∣B)⩾P(A)\left( c \right){\text{ P}}\left( {A\left| B \right.} \right) \geqslant P\left( A \right)
D) (d) None of these\left( d \right){\text{ None of these}}

Explanation

Solution

This type of question can be solved from the statement given in the question itself. And taking the lead from that we can come to the result and for this, we will take out the relation and then prove them as our need by checking the options.

Complete step by step solution:
As we have already in the question that AA and BB are two events and A⊂BA \subset B.
So, from the above condition we have
⇒A∩B=A\Rightarrow A \cap B = A
So it can also be written as,
⇒P(A∩B)=P(A)\Rightarrow P\left( {A \cap B} \right) = P\left( A \right)
Also from this, we can say that P(A)<P(B)P\left( A \right) < P\left( B \right).
Now let us consider that P(A∣B){\text{P}}\left( {A\left| B \right.} \right), and it will be equal to
⇒ P(A∣B)=P(A∩B)P(B)=P(B)P(A)≠P(A)P(B)\Rightarrow {\text{ P}}\left( {A\left| B \right.} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}} = \dfrac{{P\left( B \right)}}{{P\left( A \right)}} \ne \dfrac{{P\left( A \right)}}{{P\left( B \right)}}, let suppose it equation 11
Let us consider that P(A∣B){\text{P}}\left( {A\left| B \right.} \right), and it will be equal to
⇒ P(A∣B)=P(A∩B)P(B)=P(A)P(B)\Rightarrow {\text{ P}}\left( {A\left| B \right.} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}} = \dfrac{{P\left( A \right)}}{{P\left( B \right)}}, let suppose it equation 22
Since, according to the question, we have P(B)⩽1P\left( B \right) \leqslant 1
Therefore, from the above statement, it can be written as
⇒1P(B)⩾1\Rightarrow \dfrac{1}{{P\left( B \right)}} \geqslant 1
And it can also be written in the form of P(B) and P(A)P\left( B \right){\text{ and P}}\left( A \right), so it will be
⇒P(A)P(B)⩾P(A)\Rightarrow \dfrac{{P\left( A \right)}}{{P\left( B \right)}} \geqslant P\left( A \right)
Now from the equation 22, we get
⇒ P(A∣B)⩾P(A)\Rightarrow {\text{ P}}\left( {A\left| B \right.} \right) \geqslant P\left( A \right), let suppose it equation 33
Therefore, we can say that P(A∣B){\text{P}}\left( {A\left| B \right.} \right) is less than P(A)P\left( A \right).
Therefore, from the equation33, we can conclude that from all of the options we have only (c)\left( c \right) alternative option is correct.

Hence, the option (c)\left( c \right)is correct.

Note:
Union and Intersection are a very important concept in Probability and as well as in sets chapters. The union is spoken to by OR, though Intersection is spoken to AND. Union means taking all the common elements from both the sets once and the remaining ones also. Intersection means taking just regular components from both the sets. With an understanding of this, we can easily solve this type of problem.