A random variable X has the following probability distribution
| X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| P(X) | a | 3a | 5a | 7a | 9a | 11a | 13a | 15a | 17a |
Then the values of 'a' and \(P(0 \lt X \lt 5)\) respectively are
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{1}{81},\frac{8}{27}\)
The probabilities add up to 1:
\(a(1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17) = 1\)
The first nine odd numbers add up to \(9^2 = 81\):
\( \displaystyle 81a = 1,\quad a = \frac{1}{81} \)
\(0 \lt X \lt 5\) means X = 1, 2, 3 or 4 (the end values 0 and 5 are excluded):
\(P(0 \lt X \lt 5) = 3a + 5a + 7a + 9a = 24a\)
\( \displaystyle = \frac{24}{81} = \frac{8}{27} \)
Why the other options are incorrect
Option (2) \(\displaystyle\frac{1}{81},\frac{4}{9}\) — Incorrect. \(\frac{4}{9} = \frac{36}{81}\) is \(P(0 \le X \le 5)\), which wrongly includes X = 0 and X = 5.
Option (3) 0, 0 — Incorrect. With a = 0 all the probabilities are 0, and they cannot add up to 1.
Option (4) \(\displaystyle\frac{1}{9},\frac{4}{9}\) — Incorrect. With \(a = \frac{1}{9}\) the probabilities add up to \(\frac{81}{9} = 9\), not 1.