The probability of a specific number of successes, $X = k$, in a binomial distribution can be calculated using the following formula:

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

Where:

• $n = 13$ is the number of trials,
• $p = 0.4$ is the probability of success on each trial,
• $k = 4$ is the number of successes,
• $\binom{n}{k}$ is the binomial coefficient, which is calculated as $\frac{n!}{k!(n-k)!}$.

Now, let’s compute $P(X = 4)$:

$P(X = 4) = \binom{13}{4} (0.4)^4 (0.6)^9$

1. Calculate $\binom{13}{4}$: $\binom{13}{4} = \frac{13!}{4!(13-4)!} = \frac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1} = 715$

2. Calculate $(0.4)^4$ and $(0.6)^9$: $(0.4)^4 = 0.0256 \quad \text{and} \quad (0.6)^9 = 0.010077696$

3. Now, calculate the final probability: $P(X = 4) = 715 \times 0.0256 \times 0.010077696 = 0.1817$

Thus, the probability $P(X = 4)$ is approximately 0.1817.

Would you like further details or have any questions on this? Here are a few related questions:

1. How do you calculate the binomial coefficient $\binom{n}{k}$ for other values of $n$ and $k$?
2. How does the binomial distribution compare to the normal distribution for large $n$?
3. How can we compute cumulative probabilities for binomial distributions?
4. What does the expected value $E(X)$ represent in a binomial distribution?
5. How does changing the probability $p$ affect the shape of the binomial distribution?

Tip: In many problems, using a binomial calculator or statistical software can save time when calculating binomial probabilities.