GATE CSE 2003 | Question: 3

Question

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

• A. $\left(\dfrac{1}{4}\right),\left(\dfrac{1}{2}\right)$
• B. $\left(\dfrac{1}{2}\right),\left(\dfrac{1}{4}\right)$
• C. $\left(\dfrac{1}{2}\right),{1}$
• D.   ${1},\left(\dfrac{1}{2}\right)$

Comments

Ans

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simple explanation

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Note : If P(A)=1, then conditioning on A usually does not change probabilities, because A is already certain...So P(B|A) =P(B)

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When P(A) = 1 or P(A) = 0, A is indepdent with any event B:

when P(A) = 1:

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

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

when P(A) = 0:

$P(A \cap B) = 0$

$P(B | A) = undefined$ (because P(A) is not greater than 0)

 

NOTE: for 2 indepdendent events A and B, $P(A \cap B) = P(A)\times P(B)$

 

 

Answer 1

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. 

Comments

$P(A \cup B)$ 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.

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Here A is a surity that no matter what will happen event A definitely happens.

Now probablitiy of happening of A given B is 1 as no matter if B happens or not A is ready to happen

And happening of B given A means probability of B as A is going to happen anyway but it only to see if B is happening or not therefore probability of B

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Without explicitly mentioned A and B are independent.

There given both are independent i.e., P(A) = 1 definitely P(A') =0

So P(B) doesn't have any dependency with A so A and B are independent.

Very nice question

Answer 2

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.

Answer 3

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} $

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

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

Comments

Yes, sir did the same. 😍

Answer 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.

Option D) is the answer... 

Answer 5

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!

Answer 6

Here P(A) and P(B) are given :

So, P(A) = 1 and P(B) = ½

We have to find P(A|B) and P(B|A)

We know that P(A|B) = P(A∩B)/P(B) similarly P(B∩A)/P(A)

So we know that P(A∩B) = P(A) * P(B) = 1*½ = ½

So put this in the above formula:

P(A|B) = ½/½ = 1

P(B|A) = ½/1 = ½

So the final answer is 1 and (1/2) respectively.

Option – (D)