CUET UG 2025 — Mathematics Algebra

The probability that only X solves the problem means X successfully solves it and Y fails to solve it.

Given:
Probability that X solves the problem $= \frac{1}{3}$=31​
Probability that Y solves the problem $= \frac{1}{4}$=41​

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The probability that Y does not solve the problem:

$$P(\text{Y does not solve}) = 1 - \frac{1}{4}$$P(Y does not solve)=1−41​

$$P(\text{Y does not solve}) = \frac{3}{4}$$P(Y does not solve)=43​

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For only X to solve the problem, both events must occur: X solves AND Y does not solve.

Since these are independent events:

$$P(\text{Only X solves}) = P(\text{X solves}) \times P(\text{Y does not solve})$$P(Only X solves)=P(X solves)×P(Y does not solve)

$$P(\text{Only X solves}) = \frac{1}{3} \times \frac{3}{4}$$P(Only X solves)=31​×43​

$$P(\text{Only X solves}) = \frac{3}{12}$$P(Only X solves)=123​

$$P(\text{Only X solves}) = \frac{1}{4}$$P(Only X solves)=41​

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Therefore, the probability that only X solves the problem is $\frac{1}{4}$41​.