We are given that $X$ is a geometric random variable with $P(\text{success}) = 0.7236$. We need to find $P(X > 5 \mid X > 3)$.

This is a conditional probability, which can be expressed as:

$P(X > 5 \mid X > 3) = \frac{P(X > 5 \cap X > 3)}{P(X > 3)}$

Since $X > 5 \cap X > 3$ is equivalent to $X > 5$, the formula simplifies to:

$P(X > 5 \mid X > 3) = \frac{P(X > 5)}{P(X > 3)}$

Step 1: Finding $P(X > k)$ for geometric distribution

For a geometric random variable, the probability that $X$ is greater than $k$ is given by:

$P(X > k) = (1 - p)^k$

where $p = 0.7236$.

Step 2: Calculating $P(X > 5)$ and $P(X > 3)$

Using $p = 0.7236$:

• $P(X > 5) = (1 - 0.7236)^5 = 0.2764^5$
• $P(X > 3) = (1 - 0.7236)^3 = 0.2764^3$

Let’s calculate these values.

The conditional probability $P(X > 5 \mid X > 3)$ is approximately $0.0764$.

Do you have any further questions, or would you like more details on any part of the solution?

Here are 5 related questions for deeper understanding:

1. What is the definition of a geometric distribution and its key properties?
2. How would the result change if $P(\text{success})$ were different?
3. How can we calculate the mean and variance of a geometric distribution?
4. What is the interpretation of conditional probability in this context?
5. How would we approach the problem if we needed $P(X \leq 5 \mid X > 3)$?

Tip: In a geometric distribution, the memoryless property states that the probability of future events does not depend on past events, which simplifies conditional probabilities.