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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.   ${1},\left(\dfrac{1}{2}\right)$

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.

We can think as A is sample space.

P(A | B)= P(A)= 1

P(B| A)= P(B)= 1/2
How is P(A U B) = 1 here?
$P(A \cup B)$ means either A would occur or B would occur,

since A would definitely occur i.e. True $\implies$ $A \cup B$ would also be True.
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.

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.

An event is a subset of Universe of sample space (P(U) = 1).

Notice here, P(A) = 1, so you can thing of A as whole universe,

event B is a subset of event A.

$P(A \cap B) = P(B) = \frac{1}{2}$

$P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1$

$P(B/A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{1}{2}}{1} = \frac{1}{2}$

1 comment

Yes, sir did the same. 😍