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Let $P(E)$ denote the probability of the event $E$. Given $P(A) = 1$, $P(B) =\dfrac{1}{2}$, the values of $P(A\mid B)$ and $P(B\mid A)$ respectively are

  1. $\left(\dfrac{1}{4}\right),\left(\dfrac{1}{2}\right)$
  2. $\left(\dfrac{1}{2}\right),\left(\dfrac{1}{4}\right)$
  3. $\left(\dfrac{1}{2}\right),{1}$
  4.   ${1},\left(\dfrac{1}{2}\right)$
asked in Probability by Veteran (69k points)
edited by | 1.2k views

5 Answers

+23 votes
Best answer

It immediately follows from the monotonicity property that , 
$0\leq P(E)\leq 1,$

probability of atleast one means union of the probability of events , i.e.
$P(A\cup B) = P(A) + P(B) - P(A\cap B)$ 
here , $P(A\cup B) = 1$ , because it can not be more than $1$ and if atleast one of event have probability $1$ (here , $P(A) = 1$) , then union of both should be $1.$ 
so ,
$P(A\cup B) = P(A) + P(B) - P(A\cap B)$

    $1 = 1 + \dfrac{1}{2} - P(A\cap B) ,$
    $P(A\cap B) =\dfrac{1}{2},$ 
now ,
$P(A|B) = \dfrac{P(A\cap B)}{P(B)} = \dfrac{\left(\dfrac{1}{2}\right)}{\left(\dfrac{1}{2}\right)} = 1 ,$
$P(B|A) = \dfrac{P(A\cap B)}{P(A)} =\dfrac{\left(\dfrac{1}{2}\right)} { 1} =\dfrac{1}{2} .$ 
hence option (D) , 
NOTE :- if atleast one of the two events has/have the probability 1 . then both events should be independent events but vise - versa not true . 

answered by Boss (5.9k points)
edited by
Also, notice A and B are independent because

P(A|B) =P(A)

and P(B|A)=1/2=P(B)

By Definition of conditional probability,

P(A|B)=$\frac{P(A\cap B)}{P(B)}$

And Since P(A|B)=P(A)

this gives

P(A$\cap$B)=P(A).P(B).
@Ayush. Can you tell how to reach on conculusion

$P(A|B) =P(A)$

?

@rahul

when A and B are independent events

(1)Intuitive way: When A and B are independent, if A happens or does not happen it should not affect B, similarly if B happens or does not happens, it should not affect A.

so P(A|B)=A and P(B|A)=B.

(2)Proof :

When A and B are independent events, they can  occur together and probability of them occurring together is

P(A$\cap$B) = P(A).P(B)

and Since, $P(A|B) = \frac{P(A\cap B)}{P(B)}$

So, P(A|B)= P(A).

Similarly goes for P(B|A)=B.

Yes, but when A and B are mutually exclusive, then it means that A and B are dependent.

When A occurs, it rules out the probability of occurrence of B, and When B occurs it rules out the probability of occurrence of A.

+15 votes
P(A) = 1.

So no matter what happens with B , A always happens.
If you check options there is no need to even think in this question.

Only Option D is correct. As happening or not happening of even B does not reduce probability of Even A from 1 to 1/2 or 1/4.
answered by Veteran (49.5k points)
+3 votes

think like OSA:

Probability of A=1.

Probability of B=1/2


Probability of A given that B has happened, P(A|B)= 1
it has to be one. BECAUSE. Probability of A is 1
and It will still be 1 even if B has happened(unless B reduces it)
Since A and B(intersection) is non zero.

SO only option D is right!

answered by Loyal (3.4k points)
+2 votes

P(A/B) = 1. 

Reason : 

P(A) = 1 says that Event A will always happen irrespective of whether B happens or not.

So P(A/B) = P(A/B') = P(A) = 1.

P(B/A) = 1/2

Reason : 

      Statement - 1 : Probability of B given that A has happened = 1/2.

      Statement - 2 : A will always happen (i.e) P(A) = 1 .

      Since both Statement-1 and Statement-2 are given in question, we can always assume that both are TRUE.

If we assume both as TRUE, then conclusion we can draw is " Probability of B is always 1/2 as A will always happen".

So P(B) = P(B/A) = 1/2.

Option D) is the answer... 

 

answered by Boss (7.7k points)
+1 vote
First of all the divison must be half because p(A/B) / p(B/A)=1 . 1/2 so the answer must be option c or option d ..now p(A/B) is always 1 as the p(A)=1 and it doesnt depend any other event so option d is correct
answered by Veteran (14.3k points)
option checking ??
then P(A/B) = 1 because A always happens i.e.P(A)=1
only option is D.. :-P
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