CUET UG 2024 — Mathematics Calculus

Let's find the value of constant $c$c.

Since this is a probability distribution, the sum of all probabilities must equal 1.

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

$$0.1 + c(1) + c(2) + c(5-3) + c(5-4) = 1$$0.1+c(1)+c(2)+c(5−3)+c(5−4)=1

$$0.1 + c + 2c + 2c + c = 1$$0.1+c+2c+2c+c=1

$$0.1 + 6c = 1$$0.1+6c=1

$$6c = 0.9$$6c=0.9

$$c = 0.15$$c=0.15

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Now let's calculate each probability:

$$P(X = 0) = 0.1$$P(X=0)=0.1

$$P(X = 1) = c(1) = 0.15$$P(X=1)=c(1)=0.15

$$P(X = 2) = c(2) = 0.3$$P(X=2)=c(2)=0.3

$$P(X = 3) = c(5-3) = 0.3$$P(X=3)=c(5−3)=0.3

$$P(X = 4) = c(5-4) = 0.15$$P(X=4)=c(5−4)=0.15

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Let's find the requested probabilities:

(A) $c = 0.15$c=0.15

(B) $P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.1 + 0.15 + 0.3 = 0.55$P(X≤2)=P(X=0)+P(X=1)+P(X=2)=0.1+0.15+0.3=0.55

(C) $P(X = 2) = 0.3$P(X=2)=0.3

(D) $P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4) = 0.3 + 0.3 + 0.15 = 0.75$P(X≥2)=P(X=2)+P(X=3)+P(X=4)=0.3+0.3+0.15=0.75

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Therefore, the matching is:

(A) $c$c → (IV) 0.15

(B) $P(X \leq 2)$P(X≤2) → (III) 0.55

(C) $P(X = 2)$P(X=2) → (II) 0.3

(D) $P(X \geq 2)$P(X≥2) → (I) 0.75