CUET UG 2024 — Mathematics Algebra

First, we need to find the value of $k$k.

For any probability distribution, the sum of all probabilities must equal 1:

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

$$k + 2k + 3k = 1$$k+2k+3k=1

$$6k = 1$$6k=1

$$k = \dfrac{1}{6}$$k=61​

---

Next, let's find $P(X < 2)$P(X<2):

$$P(X < 2) = P(X = 0) + P(X = 1)$$P(X<2)=P(X=0)+P(X=1)

$$P(X < 2) = k + 2k = 3k = 3 \cdot \dfrac{1}{6} = \dfrac{1}{2}$$P(X<2)=k+2k=3k=3⋅61​=21​

---

For the expected value $E(X)$E(X):

$$E(X) = \sum[x \cdot P(X = x)]$$E(X)=∑[x⋅P(X=x)]

$$E(X) = 0 \cdot k + 1 \cdot 2k + 2 \cdot 3k$$E(X)=0⋅k+1⋅2k+2⋅3k

$$E(X) = 2k + 6k = 8k = 8 \cdot \dfrac{1}{6} = \dfrac{4}{3}$$E(X)=2k+6k=8k=8⋅61​=34​

---

For $P(1 \leq X \leq 2)$P(1≤X≤2):

$$P(1 \leq X \leq 2) = P(X = 1) + P(X = 2)$$P(1≤X≤2)=P(X=1)+P(X=2)

$$P(1 \leq X \leq 2) = 2k + 3k = 5k = 5 \cdot \dfrac{1}{6} = \dfrac{5}{6}$$P(1≤X≤2)=2k+3k=5k=5⋅61​=65​

---

Matching the lists:

(A) $k = \dfrac{1}{6}$k=61​ matches with (IV)

(B) $P(X < 2) = \dfrac{1}{2}$P(X<2)=21​ matches with (III)

(C) $E(X) = \dfrac{4}{3}$E(X)=34​ matches with (II)

(D) $P(1 \leq X \leq 2) = \dfrac{5}{6}$P(1≤X≤2)=65​ matches with (I)