CUET UG 2025 — Mathematics Algebra

Let $E_1$E1​ = die actually shows a six, $E_2$E2​ = die does not show a six, and $A$A = man reports that it is a six.

$$P(\text{truth}) = \dfrac{3}{4}$$P(truth)=43​

$$P(\text{lies}) = 1 - \dfrac{3}{4} = \dfrac{1}{4}$$P(lies)=1−43​=41​

$$P(E_1) = \dfrac{1}{6}$$P(E1​)=61​

$$P(E_2) = \dfrac{5}{6}$$P(E2​)=65​

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If the die is a six, the man reports six only when he speaks truth:

$$P(A|E_1) = \dfrac{3}{4}$$P(A∣E1​)=43​

If the die is not a six, the man reports six only when he lies:

$$P(A|E_2) = \dfrac{1}{4}$$P(A∣E2​)=41​

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Using Bayes' Theorem:

$$P(E_1|A) = \dfrac{P(A|E_1) \cdot P(E_1)}{P(A|E_1) \cdot P(E_1) + P(A|E_2) \cdot P(E_2)}$$P(E1​∣A)=P(A∣E1​)⋅P(E1​)+P(A∣E2​)⋅P(E2​)P(A∣E1​)⋅P(E1​)​

$$= \dfrac{\frac{3}{4} \times \frac{1}{6}}{\frac{3}{4} \times \frac{1}{6} + \frac{1}{4} \times \frac{5}{6}}$$=43​×61​+41​×65​43​×61​​

$$= \dfrac{\frac{3}{24}}{\frac{3}{24} + \frac{5}{24}}$$=243​+245​243​​

$$= \dfrac{\frac{3}{24}}{\frac{8}{24}}$$=248​243​​

$$= \dfrac{3}{8}$$=83​

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The probability that it is actually a six is $\boxed{\dfrac{3}{8}}$83​​