If the random variable x has the following distribution:
| \(X\) | 0 | 1 | 2 | otherwise |
| \(P(x)\) | \(k\) | \(k\) | \(2k\) | 0 |
Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | \(k\) | (I) | \(\displaystyle\frac{3}{4}\) |
| (B) | \(P(x \ge 2)\) | (II) | \(\displaystyle\frac{1}{4}\) |
| (C) | \(P(X \le 2)\) | (III) | \(\displaystyle\frac{1}{2}\) |
| (D) | \(P(0 \lt x \le 2)\) | (IV) | 1 |
Choose the correct answer from the options given below.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
X takes only the values 0, 1 and 2, with probabilities k, k and 2k.
(A) \(k\)
This matches (II) \(\frac{1}{4}\), since the probabilities add up to 1:
\(k + k + 2k = 1\)
\(\displaystyle k = \frac{1}{4}\)
(B) \(P(x \ge 2)\)
This matches (III) \(\frac{1}{2}\); the only value \(\ge 2\) is 2:
\(\displaystyle P(X = 2) = 2k = \frac{1}{2}\)
(C) \(P(X \le 2)\)
This matches (IV) 1; every value of X is \(\le 2\):
\(P(X \le 2) = k + k + 2k = 4k = 1\)
(D) \(P(0 \lt x \le 2)\)
This matches (I) \(\frac{3}{4}\); the values are 1 and 2 (0 is excluded):
\(\displaystyle P(X = 1) + P(X = 2) = k + 2k = \frac{3}{4}\)
So the matching is (A)-(II), (B)-(III), (C)-(IV), (D)-(I).