To determine the probability $P(X = 7)$ for a binomial random variable $X$ with parameters $p = 0.3$ and $n = 10$, we can use the binomial probability formula:

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

where:

• $n = 10$ is the number of trials,
• $k = 7$ is the number of successes,
• $p = 0.3$ is the probability of success on each trial.

Let's calculate it step by step.

1. Calculate the binomial coefficient $\binom{10}{7}$:

$\binom{10}{7} = \frac{10!}{7!(10-7)!} = \frac{10!}{7! \cdot 3!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$

2. Calculate the probability:

$P(X = 7) = 120 \times (0.3)^7 \times (0.7)^{3}$

Now, let's compute this.

$P(X = 7) = 120 \times (0.0002187) \times (0.343)$

$P(X = 7) \approx 120 \times 0.0000749501 = 0.00899401$

Rounding to three decimal places:

$P(X = 7) \approx 0.009$

Thus, the probability $P(X = 7)$ is approximately 0.009.

Would you like more details or have any other questions? Here are some related questions:

1. How does changing $p$ affect the binomial distribution?
2. What is the probability $P(X \leq 7)$ for the given parameters?
3. Can you explain the cumulative distribution function for a binomial variable?
4. How would the probability change if $n$ were increased to 15?
5. How can the binomial distribution be approximated by a normal distribution?

Tip: For small $p$ and large $n$, the binomial distribution can be approximated by a Poisson distribution with $\lambda = np$.