CUET UG 2025 — Mathematics Algebra

Given $P(A) = \frac{1}{2}$P(A)=21​, $P(B) = \frac{1}{3}$P(B)=31​, and $P(B|A) = \frac{1}{4}$P(B∣A)=41​

To find $P(A|B)$P(A∣B), we first need $P(A \cap B)$P(A∩B).

Using the conditional probability formula:

$$P(B|A) = \frac{P(A \cap B)}{P(A)}$$P(B∣A)=P(A)P(A∩B)​

$$\frac{1}{4} = \frac{P(A \cap B)}{\frac{1}{2}}$$41​=21​P(A∩B)​

$$P(A \cap B) = \frac{1}{4} \times \frac{1}{2}$$P(A∩B)=41​×21​

$$P(A \cap B) = \frac{1}{8}$$P(A∩B)=81​

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Now applying the conditional probability formula:

$$P(A|B) = \frac{P(A \cap B)}{P(B)}$$P(A∣B)=P(B)P(A∩B)​

$$P(A|B) = \frac{\frac{1}{8}}{\frac{1}{3}}$$P(A∣B)=31​81​​

$$P(A|B) = \frac{1}{8} \times \frac{3}{1}$$P(A∣B)=81​×13​

$$P(A|B) = \frac{3}{8}$$P(A∣B)=83​

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Therefore, $P(A|B) = \frac{3}{8}$P(A∣B)=83​