ISRO-2013-71

Question

Let $P(E)$ denote the probability of the occurrence of event $E$. If $P(A)= 0.5$ and $P(B)=1$ then the values of $P(A|B)$ and $P(B|A)$ respectively are

• A. $0.5, 0.25$
• B. $0.25, 0.5$
• C. $0.5, 1$
• D. $1, 0.5$

Comments

Since P(B)=1, event B denotes the sample space and event A where P(A)<1, denotes an event...

Hence A intersection B = A

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i quickly drawed tree diagram and got to know that PR(B|A) is 1,

only option C satisfies this,

also, we can  calculate PR(A|B) but we can save time in exam just by looking at option

Answer 1

P(A)= 0.5

P(B)=1

Here there is no dependency in event A and B. 
So  P(A $\cap$ B) = P(A) * P(B)

P(A/B)= probability of occurrence  of event A when B has already occurred

            = P(A $\cap$ B) / P(B) 

            = (0.5 * 1) /1 = 0.5

P(B/A)= probability of occurrence  of event B when A  has already occurred 

            = P(B $\cap$ A) / P(A) 

            = (1 * 0.5) / 0.5 =  1

Ans- C

         

Comments

 @Manoja Rajalakshmi A

C also satisfiy your equation then how could you say that only B is the option...if they are dependent then ...

P(B) would be less that 1...

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Thank you.Can you please clear it again?

I am not saying option B is answer. I am asking if A and B are dependent (say given in question) then can option B be the answer(and not C)? This is my doubt.

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NOTE. IF SOMEONE THINKS LIKE FOLLOWING WAYS, THEN HE/SHE IS LACKING CONCEPT OF PROBABILITY.

BY FOLLOWING WAYS, ANSWER IS CORRECT, BUT METHOD IS ABSOLUTELY WRONG.

 

We know from set theory A INTERSECTION B =  A-(A-B)

Here, SET B CONTAINS MORE ELEMENT THAN SET A as P(B)=1 and P(A)=0.5

Therefore, A INTERSECTION B = B-(B-A) = 1-(1-0.5)= 1-0.5= 0.5

now, substituting value of A INTERSECTION B in respective equations we can get the answers. using

P(B/A)= P(A INTERSECTION B)/P(A) as A INTERSECTION B = B INTERSECTION A ( from set theory property)

Hence, option C is the correct option.