NIMCET 2012 — Mathematics PYQ

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Question

NIMCET 2012 — Mathematics PYQ

NIMCET | Mathematics | 2012

Let $P(E)$ denote the probability of event $E$ . Given $P(A) = 1$ and $P(B) = \frac{1}{2}$ , the values of $P(A|B)$ and $P(B|A)$ respectively are:

Choose the correct answer:

Explanation

1. Finding $P(A \cap B)$ :

Since $P(A) = 1$ , event $A$ is a sure event (it covers the entire sample space). Therefore, the intersection of any event $B$ with a sure event $A$ is simply the event $B$ itself.

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

2. Calculating $P(A|B)$ :

Using the formula for conditional probability:

P(A|B) = \frac{P(A \cap B)}{P(B)}

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

P(A|B) = 1

3. Calculating $P(B|A)$ :

Using the formula for conditional probability:

P(B|A) = \frac{P(A \cap B)}{P(A)}

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

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

Conclusion:

The values are $1$ and $\frac{1}{2}$ .

Correct Option: (d)