If \(X\) is a random variable which can assume values 0, 1, 2, 3 or 4 such that \(P(X = 1) = P(X = 2)\) and \(3P(X = 3) = 4P(X = 4) = P(X = 0) = \displaystyle\frac{1}{8}\) then \(P(X \gt 0)\) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{7}{8}\)
X takes only the values 0, 1, 2, 3 and 4, so the event X > 0 is the complement of X = 0:
\( \displaystyle P(X \gt 0) = 1 - P(X = 0) = 1 - \frac{1}{8} = \frac{7}{8} \)
The other conditions are not needed. As a check, they give a valid distribution:
\( \displaystyle P(X = 3) = \frac{1}{24},\quad P(X = 4) = \frac{1}{32} \)
\( \displaystyle 2P(X = 1) = 1 - \frac{1}{8} - \frac{1}{24} - \frac{1}{32} = \frac{77}{96} \)
so \(P(X = 1) = P(X = 2) = \frac{77}{192}\), and all the probabilities are positive.