The probability distribution of a random variable X is given by
| X | 0 | 1 | 2 |
| P(X) | \(1-7a^2\) | \(\displaystyle\frac{1}{2}a+\frac{1}{4}\) | \(a^2\) |
If \(a \gt 0\), then \(P(0 \lt X \le 2)\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{7}{16}\)
The probabilities of a distribution add up to 1:
\( \displaystyle (1 - 7a^2) + \left(\frac{1}{2}a + \frac{1}{4}\right) + a^2 = 1 \)
\( \displaystyle -6a^2 + \frac{1}{2}a + \frac{1}{4} = 0 \)
Multiplying by −4:
\(24a^2 - 2a - 1 = 0\)
\((4a - 1)(6a + 1) = 0\)
Since a > 0, \(a = \frac{1}{4}\).
The event 0 < X ≤ 2 means X = 1 or X = 2 (X = 0 is excluded):
\( \displaystyle P(X = 1) = \frac{1}{8} + \frac{1}{4} = \frac{3}{8} \)
\( \displaystyle P(X = 2) = \frac{1}{16} \)
\( \displaystyle P(0 \lt X \le 2) = \frac{3}{8} + \frac{1}{16} = \frac{7}{16} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{1}{16}\) — Incorrect. This is P(X = 2) alone; P(X = 1) must be added.
Option (2) \(\displaystyle\frac{3}{18}\) — Incorrect. The required probability is \(\frac{7}{16}\), not \(\frac{3}{18}\).
Option (4) \(\displaystyle\frac{9}{16}\) — Incorrect. This is \(P(X = 0) = 1 - \frac{7}{16}\), the probability of the excluded value.