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

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

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33 votes

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

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

45 votes

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. **

5 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... **

4 votes

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} $

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

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