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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.   ${1},\left(\dfrac{1}{2}\right)$
edited | 1.5k views

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

Probability of at least 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 at least one of the event has 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,
$P(A\mid B) = \dfrac{P(A\cap B)}{P(B)} = \dfrac{\left(\dfrac{1}{2}\right)}{\left(\dfrac{1}{2}\right)} = 1 ,$
$P(B\mid A) = \dfrac{P(A\cap B)}{P(A)} =\dfrac{\left(\dfrac{1}{2}\right)} { 1} =\dfrac{1}{2} .$
Hence, option is $(D)$.

NOTE :- if at least one of the two events has probability 1, then both events should be independent but vise versa is not true.

edited
+2
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).
0
@Ayush. Can you tell how to reach on conculusion

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

?
+1

@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.

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.

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!

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.

+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
+3
option checking ??
then P(A/B) = 1 because A always happens i.e.P(A)=1
only option is D.. :-P

An analogy that may help to relate with the question :

Say we are living in an area where it rains everyday (Event A) and there are two groups of day MWF and TTS (Just assume that Sunday doesn't exist at all). Let Event B be that the chosen day is one amongst MWF. Clearly the event A has probability of 1 and probability of event B is 1/2.

Now the question can be interpreted as :

P(A|B) = Given that the day chosen is among MWF, probability of raining ?

Since it is raining no matter what, therefore the probability of this part is 1.

P(B|A) = Given that it is raining, probability that the day is one of MWF ?

Since it rains on all days, the probability that the randomly chosen day is one amongst MWF is 3/6 or 1/2.