CUET UG 2025 — Mathematics Algebra

Since this is a valid probability distribution, all probabilities must sum to 1.

$$\frac{25k}{72} + \frac{15k}{72} + \frac{3k}{72} + \frac{k}{360} = 1$$7225k​+7215k​+723k​+360k​=1

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Taking $k$k common:

$$k\left(\frac{25}{72} + \frac{15}{72} + \frac{3}{72} + \frac{1}{360}\right) = 1$$k(7225​+7215​+723​+3601​)=1

LCM of 72 and 360 is 360:

$$\frac{25}{72} = \frac{125}{360}, \quad \frac{15}{72} = \frac{75}{360}, \quad \frac{3}{72} = \frac{15}{360}, \quad \frac{1}{360} = \frac{1}{360}$$7225​=360125​,7215​=36075​,723​=36015​,3601​=3601​

$$k \times \frac{125 + 75 + 15 + 1}{360} = 1$$k×360125+75+15+1​=1

$$k \times \frac{216}{360} = 1$$k×360216​=1

$$k \times \frac{3}{5} = 1$$k×53​=1

$$k = \dfrac{5}{3}$$k=35​

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Here $m = 5$m=5 and $n = 3$n=3, with $\gcd(5, 3) = 1$gcd(5,3)=1.

$$m + n = 5 + 3 = 8$$m+n=5+3=8