Which of the following can be a probability distribution function for a random variable X?
A.
| X | 0 | 1 | 2 |
| P(x) | 0.4 | 0.4 | 0.2 |
B.
| X | 1 | 2 | 3 | 4 |
| P(x) | 0.1 | 0.4 | 0.3 | −0.2 |
C.
| X | 1 | 2 | 3 | 4 |
| P(x) | \(\displaystyle\frac{1}{5}\) | \(\displaystyle\frac{2}{5}\) | \(\displaystyle\frac{1}{5}\) | \(\displaystyle\frac{1}{5}\) |
D.
| X | 0 | 1 | 2 | 3 |
| P(x) | 0.2 | 0.3 | 0.2 | 0.3 |
E.
| X | 1 | 2 | 3 |
| P(x) | \(\displaystyle\frac{1}{3}\) | \(\displaystyle\frac{2}{3}\) | \(\displaystyle\frac{1}{3}\) |
Choose the correct answer from the options given below
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A, C, D only
A table is a probability distribution only if every probability lies between 0 and 1 and the probabilities add up to 1. The sums:
\( \text{A: } 0.4 + 0.4 + 0.2 = 1 \)
\( \displaystyle \text{C: } \frac{1}{5} + \frac{2}{5} + \frac{1}{5} + \frac{1}{5} = \frac{5}{5} = 1 \)
\( \text{D: } 0.2 + 0.3 + 0.2 + 0.3 = 1 \)
\( \displaystyle \text{E: } \frac{1}{3} + \frac{2}{3} + \frac{1}{3} = \frac{4}{3} \)
So A, C and D have non-negative probabilities adding up to 1; B has a negative value and E adds up to \(\frac{4}{3}\).
Checking the statements
(A) X: 0, 1, 2 with P(x): 0.4, 0.4, 0.2 — Correct. All values are non-negative and add up to 1.
(B) X: 1, 2, 3, 4 with P(x): 0.1, 0.4, 0.3, −0.2 — Incorrect. A probability cannot be negative (−0.2).
(C) X: 1, 2, 3, 4 with P(x): \(\frac{1}{5}, \frac{2}{5}, \frac{1}{5}, \frac{1}{5}\) — Correct. All values are non-negative and add up to 1.
(D) X: 0, 1, 2, 3 with P(x): 0.2, 0.3, 0.2, 0.3 — Correct. All values are non-negative and add up to 1.
(E) X: 1, 2, 3 with P(x): \(\frac{1}{3}, \frac{2}{3}, \frac{1}{3}\) — Incorrect. The probabilities add up to \(\frac{4}{3} \gt 1\).