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 $P(X > 0)$ is

To solve this, we first need to identify the possible sums when throwing two dice and determine which are perfect squares.

When two dice are thrown, the sum $S$ can range from $2$ to $12$. The perfect squares in this range are:

• $4$ (combinations: $(1,3), (2,2), (3,1)$) — 3 ways
• $9$ (combinations: $(3,6), (4,5), (5,4), (6,3)$) — 4 ways

Total number of ways to get a perfect square sum = $3 + 4 = 7$.

Total possible outcomes for two dice = $36$.

Therefore:

• Probability of a perfect square sum $P(S) = \frac{7}{36}$
• Probability of a non-perfect square sum $P(S') = 1 - \frac{7}{36} = \frac{29}{36}$

Understand the Random Variable $X$:

$X$ is the number of non-perfect square sums that appear before the experiment stops. The experiment stops if:

1. A perfect square sum appears.
2. Non-perfect square sums appear 5 times in a row.

Calculate $P(X > 0)$:

$P(X > 0)$ is the probability that at least one non-perfect square sum appears.

It is easier to calculate this using the complement:

$P(X > 0) = 1 - P(X = 0)$

$X = 0$ means that zero non-perfect square sums appeared. This only happens if a perfect square sum is thrown on the very first attempt.

$P(X = 0) = P(\text{perfect square on 1st throw}) = \frac{7}{36}$

Now, substitute this back into the complement formula:

$P(X > 0) = 1 - \frac{7}{36}$

$P(X > 0) = \frac{29}{36}$