Given the problem, we are dealing with a geometric distribution where the probability of encountering a bug on each run is $p = 0.15$, and the random variable $X$ represents the number of runs until the first bug is found. The probability mass function (PMF) of a geometric random variable is:

$P(X = k) = (1 - p)^{k-1} p$

Now, we are given that the program has already been run 3 times without encountering a bug. We need to find the probability that the first bug will be encountered after the 5th run, which means that the first bug is found on the 6th run or later.

Step 1: Understand the conditional probability

We are interested in the probability that the first bug is found after the 5th run, given that the first 3 runs did not encounter any bugs. This is a conditional probability problem. The conditional probability is:

$P(X > 5 \mid X > 3)$

Using the memoryless property of the geometric distribution, the probability that the first bug is found after the 5th run, given that no bug was found in the first 3 runs, is the same as the probability that the first bug is found after the 2nd run (since the process "restarts" after the first 3 runs). Therefore:

$P(X > 5 \mid X > 3) = P(X > 2)$

Step 2: Calculate $P(X > 2)$

The probability that no bug is found in the first 2 runs is:

$P(X > 2) = (1 - p)^2 = (1 - 0.15)^2 = 0.85^2 = 0.7225$

Step 3: Final answer

Thus, the probability that the first bug will be found after the 5th run, given that the program has already been run 3 times without finding a bug, is 0.72 (rounded to two decimal places).

Summary:

$P(X > 5 \mid X > 3) = 0.72$

Would you like more details or have any further questions? Here are some related questions to explore:

1. How do we calculate the probability for other values of $X$ in a geometric distribution?
2. What is the expected number of runs until the first bug is found?
3. How does the memoryless property apply to different types of distributions?
4. How would the probability change if the chance of finding a bug on each run were higher (say, 30%)?
5. What other distributions are memoryless, and how do they compare to the geometric distribution?

Tip: The memoryless property of the geometric distribution means that the past doesn't affect the future, which simplifies many problems like this one.