To solve for $P(X \leq 2)$ for a binomial distribution with parameters $n = 6$ trials and $p = 0.3$ as the probability of success, we can use the binomial probability formula:

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

• $n = 6$ is the number of trials,
• $p = 0.3$ is the probability of success,
• $k$ is the number of successes,
• $\binom{n}{k}$ is the binomial coefficient.

To find $P(X \leq 2)$, we need to calculate:

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

We'll compute each term:

1. For $P(X = 0)$: $P(X = 0) = \binom{6}{0} (0.3)^0 (0.7)^6 = 1 \cdot 1 \cdot (0.7)^6 = 0.117649$

2. For $P(X = 1)$: $P(X = 1) = \binom{6}{1} (0.3)^1 (0.7)^5 = 6 \cdot 0.3 \cdot (0.7)^5 = 6 \cdot 0.3 \cdot 0.16807 = 0.302526$

3. For $P(X = 2)$: $P(X = 2) = \binom{6}{2} (0.3)^2 (0.7)^4 = 15 \cdot 0.09 \cdot (0.7)^4 = 15 \cdot 0.09 \cdot 0.2401 = 0.324135$

Now, summing these probabilities:

$P(X \leq 2) = 0.117649 + 0.302526 + 0.324135 = 0.744310$

Thus, $P(X \leq 2) \approx 0.7443$.

Answer: $P(X \leq 2) \approx 0.7443$

Let me know if you would like further details or explanations on any part!

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Further questions for you:

1. How does the binomial distribution behave with different values of $p$?
2. What would the probability be if the number of trials were increased to $n = 10$?
3. How can the cumulative binomial probability be calculated using a statistical table?
4. Can the binomial distribution approximate a normal distribution in any cases? If so, when?
5. How does the standard deviation of a binomial distribution change with varying $n$ and $p$?

Tip: When calculating binomial probabilities by hand, make sure to carefully compute the binomial coefficients, as small errors can accumulate. Using a calculator or statistical software is a great way to ensure accuracy!