A pair of dice is thrown until the sum of numbers appeared is a perfect square or a non-perfect square sum appeared five times in succession. If random variable X denotes the number of non perfect square sums appeared then \(P(X \gt 0)\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{29}{36}\)
X = 0 exactly when the first throw already gives a perfect square sum, so by the complement
\(P(X \gt 0) = 1 - P(X = 0)\)
The sum of two dice lies between 2 and 12, so the only perfect squares possible are 4 and 9:
Sum 4: (1, 3), (2, 2), (3, 1), 3 outcomes
Sum 9: (3, 6), (4, 5), (5, 4), (6, 3), 4 outcomes
\( \displaystyle P(X = 0) = \frac{3 + 4}{36} = \frac{7}{36} \)
\( \displaystyle P(X \gt 0) = 1 - \frac{7}{36} = \frac{29}{36} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{7}{36}\) — Incorrect. This is P(X = 0), the complement of the event asked.
Option (3) 1 — Incorrect. X = 0 is possible (a perfect square on the first throw), so \(P(X \gt 0) \lt 1\).
Option (4) \(\displaystyle\frac{1}{2}\) — Incorrect. The value is \(\frac{29}{36}\), not \(\frac{18}{36}\).