CUET UG 2025 — Mathematics Algebra

In any probability distribution, the sum of all probabilities must equal 1.

$$P(X=0) + P(X=1) + P(X=2) + ... + P(X=8) = 1$$P(X=0)+P(X=1)+P(X=2)+...+P(X=8)=1

Substituting the values from the table:

$$a + 3a + 5a + 7a + 9a + 11a + 13a + 15a + 17a = 1$$a+3a+5a+7a+9a+11a+13a+15a+17a=1

$$81a = 1$$81a=1

$$a = \dfrac{1}{81}$$a=811​

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The notation $0 < X < 5$0<X<5 means $X$X is greater than 0 and less than 5.

$X$X can take values: $1, 2, 3, 4$1,2,3,4

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

$$P(0 < X < 5) = 3a + 5a + 7a + 9a$$P(0<X<5)=3a+5a+7a+9a

$$P(0 < X < 5) = 24a$$P(0<X<5)=24a

Substituting $a = \dfrac{1}{81}$a=811​:

$$P(0 < X < 5) = 24 \times \dfrac{1}{81}$$P(0<X<5)=24×811​

$$P(0 < X < 5) = \dfrac{24}{81}$$P(0<X<5)=8124​

$$P(0 < X < 5) = \dfrac{8}{27}$$P(0<X<5)=278​

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Therefore:

$a = \dfrac{1}{81}$a=811​ and $P(0 < X < 5) = \dfrac{8}{27}$P(0<X<5)=278​