CUET UG 2025 — Mathematics Algebra

The probability distribution has an unknown value $k$k. The sum of all probabilities must equal 1.

$$P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7) = 1$$P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)+P(X=7)=1

$$0 + k + 2k + 2k + 3k + k^2 + 2k^2 + 7k^2 + k = 1$$0+k+2k+2k+3k+k2+2k2+7k2+k=1

Combining like terms:

$$10k^2 + 9k = 1$$10k2+9k=1

$$10k^2 + 9k - 1 = 0$$10k2+9k−1=0

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Using the quadratic formula where $a = 10$a=10, $b = 9$b=9, $c = -1$c=−1:

$$k = \dfrac{-9 \pm \sqrt{81 - 4(10)(-1)}}{2(10)}$$k=2(10)−9±81−4(10)(−1)​​

$$k = \dfrac{-9 \pm \sqrt{81 + 40}}{20}$$k=20−9±81+40​​

$$k = \dfrac{-9 \pm \sqrt{121}}{20}$$k=20−9±121​​

$$k = \dfrac{-9 \pm 11}{20}$$k=20−9±11​

This gives $k = \dfrac{2}{20} = \dfrac{1}{10}$k=202​=101​ or $k = \dfrac{-20}{20} = -1$k=20−20​=−1

Since probability cannot be negative, $k = \dfrac{1}{10}$k=101​.

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The notation $4 < X < 7$4<X<7 means $X$X is strictly between 4 and 7, so $X$X can only be 5 or 6.

$$P(4 < X < 7) = P(X=5) + P(X=6)$$P(4<X<7)=P(X=5)+P(X=6)

$$P(4 < X < 7) = k^2 + 2k^2$$P(4<X<7)=k2+2k2

$$P(4 < X < 7) = 3k^2$$P(4<X<7)=3k2

$$P(4 < X < 7) = 3 \times \left(\dfrac{1}{10}\right)^2$$P(4<X<7)=3×(101​)2

$$P(4 < X < 7) = 3 \times \dfrac{1}{100}$$P(4<X<7)=3×1001​

$$P(4 < X < 7) = \dfrac{3}{100}$$P(4<X<7)=1003​