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).

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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

- $\left(\dfrac{1}{4}\right),\left(\dfrac{1}{2}\right)$
- $\left(\dfrac{1}{2}\right),\left(\dfrac{1}{4}\right)$
- $\left(\dfrac{1}{2}\right),{1}$
- ${1},\left(\dfrac{1}{2}\right)$

Best answer

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. **

@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.

@Ayush Upadhyaya for P(A|B) we can directly say P(A) as it is sure event. But to calculate P(B|A), we will hae to find their intersection and prove it right? We can't conclude it directly.

**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... **

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.

In a class there are 100 strudents(50 girls and 50 boys)

the teacher selects a monitor for the class.

A=event that a student is choosen

P(A)=1

B=event that a boy is choosen

P(B)=1/2.

now,P(A/B)=the probabiilty that a student is chosen given a boy is choosen is always 1.

P(B/A)=the probabiilty that a boy is chosen given a student is choosen is always 1/2.(As it can be a boy or a girl).

Answer is D

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} $

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

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

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