The random variable \( X \) has a probability distribution \( P(X) \) of the following form, where \( k \) is some real number.
\( P(X=x)=\begin{cases}k, & \text{if}\ x=0\\ \displaystyle\frac{k}{2}, & \text{if}\ x=1\ \text{or}\ 2\\0, & \text{otherwise}\end{cases} \)
On the basis of the above information, match List-I with List-II
| List-I | List-II | ||
| (A) | Value of \( k \) | (I) | \( \displaystyle\frac{1}{4} \) |
| (B) | P(X < 2) | (II) | 0 |
| (C) | P(X > 1) | (III) | \( \displaystyle\frac{1}{2} \) |
| (D) | \( P(X\gt 2) \) | (IV) | \( \displaystyle\frac{3}{4} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
(A) Value of \( k \)
This matches (III) \( \frac{1}{2} \).
The probabilities must add up to 1:
\( \displaystyle k+\frac{k}{2}+\frac{k}{2}=1 \)
\( \displaystyle 2k=1 \)
\( \displaystyle k=\frac{1}{2} \)
So P(X = 0) = 1/2, P(X = 1) = 1/4 and P(X = 2) = 1/4.
(B) P(X < 2)
This matches (IV) \( \frac{3}{4} \).
\( \displaystyle P(X\lt 2)=P(X=0)+P(X=1) \)
\( \displaystyle =\frac{1}{2}+\frac{1}{4}=\frac{3}{4} \)
(C) P(X > 1)
This matches (I) \( \frac{1}{4} \).
\( \displaystyle P(X\gt 1)=P(X=2)=\frac{1}{4} \)
(D) \( P(X\gt 2) \)
This matches (II) 0.
X takes no value greater than 2, so the probability is 0.
So the matching is (A)-(III), (B)-(IV), (C)-(I), (D)-(II).