A random variable X has the following probability distribution:
| X | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| P(X) | k | 2k | 2k | 3k | \(k^2\) | \(2k^2\) | \(7k^2 + k\) |
Match the options of List-I to List-II:
| List-I | List-II | ||
| (A) | k | (I) | \(\displaystyle\frac{7}{10}\) |
| (B) | \(P(X \lt 3)\) | (II) | \(\displaystyle\frac{53}{100}\) |
| (C) | \(P(X \gt 2)\) | (III) | \(\displaystyle\frac{1}{10}\) |
| (D) | \(P(2 \lt X \lt 7)\) | (IV) | \(\displaystyle\frac{3}{10}\) |
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)
(A) k
This matches (III) \(\frac{1}{10}\).
The probabilities add up to 1:
\(k + 2k + 2k + 3k + k^2 + 2k^2 + 7k^2 + k = 1\)
\(10k^2 + 9k - 1 = 0\)
\((10k - 1)(k + 1) = 0\)
A probability cannot be negative, so k = −1 is rejected:
\( \displaystyle k = \frac{1}{10} \)
(B) \(P(X \lt 3)\)
This matches (IV) \(\frac{3}{10}\).
\( \displaystyle P(X \lt 3) = P(1) + P(2) = 3k = \frac{3}{10} \)
(C) \(P(X \gt 2)\)
This matches (I) \(\frac{7}{10}\).
X > 2 is the complement of X < 3:
\( \displaystyle P(X \gt 2) = 1 - \frac{3}{10} = \frac{7}{10} \)
(D) \(P(2 \lt X \lt 7)\)
This matches (II) \(\frac{53}{100}\).
2 < X < 7 means X = 3, 4, 5 or 6:
\(P(2 \lt X \lt 7) = 2k + 3k + k^2 + 2k^2 = 5k + 3k^2\)
\( \displaystyle = \frac{5}{10} + \frac{3}{100} = \frac{53}{100} \)
So the matching is (A) - (III), (B) - (IV), (C) - (I), (D) - (II).