CUET UG 2025 — Mathematics Algebra

For any probability distribution, all probabilities must add up to 1.

$$P(0) + P(1) + P(2) + P(3) = 1$$P(0)+P(1)+P(2)+P(3)=1

$$\frac{k}{3} + \frac{k}{2} + \frac{k}{4} + \frac{k}{7} = 1$$3k​+2k​+4k​+7k​=1

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To add these fractions, find the LCM of the denominators 3, 2, 4, and 7.

LCM of 3, 2, 4, 7 = 84

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Convert each fraction to have denominator 84:

$$\frac{k}{3} = \frac{28k}{84}$$3k​=8428k​

$$\frac{k}{2} = \frac{42k}{84}$$2k​=8442k​

$$\frac{k}{4} = \frac{21k}{84}$$4k​=8421k​

$$\frac{k}{7} = \frac{12k}{84}$$7k​=8412k​

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$$\frac{28k + 42k + 21k + 12k}{84} = 1$$8428k+42k+21k+12k​=1

$$\frac{103k}{84} = 1$$84103k​=1

$$103k = 84$$103k=84

$$k = \frac{84}{103}$$k=10384​

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Therefore, the value of $k$k is $\frac{84}{103}$10384​.