CUET UG 2025 — Mathematics Algebra

Events A and B are independent with $P(A) = 0.3$P(A)=0.3 and $P(B) = 0.6$P(B)=0.6.

For independent events, $P(A \text{ and } B) = P(A) \times P(B)$P(A and B)=P(A)×P(B).

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(A) $P(A \text{ and } B) = P(A) \times P(B)$P(A and B)=P(A)×P(B)

$$= 0.3 \times 0.6$$=0.3×0.6

$$= 0.18$$=0.18

(A) matches with (II)

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(B) $P(\text{not } B) = 1 - P(B)$P(not B)=1−P(B)

$$= 1 - 0.6$$=1−0.6

$$= 0.4$$=0.4

$$P(A \text{ and not } B) = P(A) \times P(\text{not } B)$$P(A and not B)=P(A)×P(not B)

$$= 0.3 \times 0.4$$=0.3×0.4

$$= 0.12$$=0.12

(B) matches with (III)

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(C) $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$P(A or B)=P(A)+P(B)−P(A and B)

$$= 0.3 + 0.6 - 0.18$$=0.3+0.6−0.18

$$= 0.72$$=0.72

(C) matches with (IV)

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(D) $P(\text{not } A) = 1 - P(A)$P(not A)=1−P(A)

$$= 1 - 0.3$$=1−0.3

$$= 0.7$$=0.7

$$P(\text{neither A nor B}) = P(\text{not } A) \times P(\text{not } B)$$P(neither A nor B)=P(not A)×P(not B)

$$= 0.7 \times 0.4$$=0.7×0.4

$$= 0.28$$=0.28

(D) matches with (I)

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Final matching:

• (A) → (II) → 0.18
• (B) → (III) → 0.12
• (C) → (IV) → 0.72
• (D) → (I) → 0.28