This problem is asking us to calculate the probability of exactly $k = 1$ successes in a binomial distribution with $n = 13$ trials and a probability of success $p = 0.22$ on each trial. We can use the binomial probability formula:

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

Where:

• $n = 13$ (number of trials),
• $k = 1$ (number of successes),
• $p = 0.22$ (probability of success on a single trial),
• $\binom{n}{k}$ is the binomial coefficient, which represents the number of ways to choose $k$ successes out of $n$ trials, and is computed as $\frac{n!}{k!(n-k)!}$.

Let's calculate the value step by step:

1. Binomial coefficient:

$\binom{13}{1} = \frac{13!}{1!(13-1)!} = 13$

2. Compute $p^k$ and $(1 - p)^{n-k}$:

$p^1 = 0.22$ $(1 - p)^{12} = (0.78)^{12}$

3. Putting it all together:

$P(X = 1) = 13 \times 0.22 \times (0.78)^{12}$

I will now calculate this probability.

The probability of getting exactly $k = 1$ success in 13 trials, with a probability of success on a single trial of $p = 0.22$, is approximately:

$P(X = 1) = 0.1450$

Would you like any further details or have any questions about the solution?

Here are 5 related questions to explore:

1. How does the probability change if the number of trials $n$ increases?
2. What is the probability of getting exactly 0 successes for the same scenario?
3. How can we calculate the cumulative probability of getting at most 2 successes?
4. What is the mean and variance of this binomial distribution?
5. How would the result change if the probability of success $p$ were higher?

Tip: The binomial coefficient $\binom{n}{k}$ represents the number of different ways $k$ successes can occur out of $n$ trials, and it scales the probability calculation based on these combinations.