If the rv X has a geometric distribution with P(success)=0.7236, then find  P[(X>5)|(X>3)].

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.

If the rv X has a geometric distribution with P(success)=0.7236, then find P[(X>5)|(X>3)].

Learn how to solve a conditional probability problem involving a geometric distribution with P(success) = 0.7236. This solution covers key concepts like geometric distribution and conditional probability. Suitable for students in Grades 10-12.

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