The probability distribution of a random variable \(x\) is, \(P(x) =\displaystyle\frac{k}{2^x}, x= 0,1,2,3\). Then
Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | \(k\) | (I) | \(\displaystyle\frac{2}{15}\) |
| (B) | \(P(x=1)\) | (II) | \(\displaystyle\frac{1}{5}\) |
| (C) | \(P(1 \lt x \lt 3)\) | (III) | \(\displaystyle\frac{8}{15}\) |
| (D) | \(P(x \ge 2)\) | (IV) | \(\displaystyle\frac{4}{15}\) |
Choose the correct answer from the options given below.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
The probabilities of all values add up to 1, and this fixes k.
(A) \(k\)
This matches (III) \(\frac{8}{15}\).
\( \displaystyle \frac{k}{1} + \frac{k}{2} + \frac{k}{4} + \frac{k}{8} = 1 \)
\( \displaystyle \frac{15k}{8} = 1 \)
\( \displaystyle k = \frac{8}{15} \)
(B) \(P(x=1)\)
This matches (IV) \(\frac{4}{15}\).
\( \displaystyle P(x = 1) = \frac{k}{2} = \frac{4}{15} \)
(C) \(P(1 \lt x \lt 3)\)
This matches (I) \(\frac{2}{15}\). The only value strictly between 1 and 3 is x = 2.
\( \displaystyle P(x = 2) = \frac{k}{4} = \frac{2}{15} \)
(D) \(P(x \ge 2)\)
This matches (II) \(\frac{1}{5}\).
\( \displaystyle P(x \ge 2) = P(2) + P(3) = \frac{k}{4} + \frac{k}{8} \)
\( \displaystyle = \frac{2}{15} + \frac{1}{15} = \frac{3}{15} = \frac{1}{5} \)
So the matching is (A)-(III), (B)-(IV), (C)-(I), (D)-(II).