To find $P(X = 7)$ for a binomial distribution $X \sim \text{Bin}(9, 0.2)$, we use the binomial probability formula:

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

Where:

• $n = 9$ (the number of trials),
• $p = 0.2$ (the probability of success),
• $k = 7$ (the number of successes).

Step 1: Calculate the binomial coefficient $\binom{9}{7}$

$\binom{9}{7} = \frac{9!}{7!(9-7)!} = \frac{9!}{7!2!} = \frac{9 \times 8}{2 \times 1} = 36$

Step 2: Calculate the probability

Now, we apply the binomial formula: $P(X = 7) = 36 \times (0.2)^7 \times (0.8)^2$

We compute each part:

• $(0.2)^7 = 0.000128$
• $(0.8)^2 = 0.64$

$P(X = 7) = 36 \times 0.000128 \times 0.64 = 36 \times 0.00008192 = 0.00294432$

Thus, rounding to four decimal places: $P(X = 7) \approx 0.0029$

Final Answer:

$P(X = 7) \approx 0.0029$

Would you like a detailed breakdown or have any questions about this solution?

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Here are 5 related questions you might find interesting:

1. How does the binomial distribution formula change when the probability $p$ increases to 0.5?
2. What would be the probability for $P(X = 0)$ in this same binomial distribution?
3. How do you find the expected value and variance of a binomial distribution?
4. How can you calculate cumulative probabilities in a binomial distribution, such as $P(X \leq 7)$?
5. What is the difference between the binomial and normal distributions, and when can you use the normal approximation?

Tip: When working with binomial probabilities, it's helpful to check for conditions where you can use the normal approximation for easier calculations, especially when $n$ is large.