CUET UG 2022 — Mathematics Algebra

Independent events with $P(A)=\frac{1}{3}, P(B)=\frac{1}{4}, P(C)=\frac{1}{5}$P(A)=31​,P(B)=41​,P(C)=51​.

$$P(\text{exactly one}) = P(A)P(B')P(C') + P(A')P(B)P(C') + P(A')P(B')P(C)$$P(exactly one)=P(A)P(B′)P(C′)+P(A′)P(B)P(C′)+P(A′)P(B′)P(C)
$= \frac{1}{3}\cdot\frac{3}{4}\cdot\frac{4}{5} + \frac{2}{3}\cdot\frac{1}{4}\cdot\frac{4}{5} + \frac{2}{3}\cdot\frac{3}{4}\cdot\frac{1}{5} = \frac{12+8+6}{60} = \frac{26}{60} = \frac{13}{30}$=31​⋅43​⋅54​+32​⋅41​⋅54​+32​⋅43​⋅51​=6012+8+6​=6026​=3013​.

Note: The computed value is $\frac{13}{30}$3013​. Option (4) appears as $\frac{13}{20}$2013​, which we read as a print error for $\frac{13}{30}$3013​, matching the official key.