To solve this problem, we use the Binomial distribution formula:

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

where:

• $n = 13$ is the number of trials,
• $p = 0.25$ is the probability of rolling a "2" on a single roll,
• $k$ is the number of successes (number of times "2" is rolled),
• $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ is the number of combinations.

The probability of rolling a "2" at most 2 times is:

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

Step 1: Compute individual probabilities

For $X = 0$:

$P(X = 0) = \binom{13}{0} (0.25)^0 (0.75)^{13} = 1 \cdot 1 \cdot (0.75)^{13}.$

$P(X = 0) = (0.75)^{13} \approx 0.01336.$

For $X = 1$:

$P(X = 1) = \binom{13}{1} (0.25)^1 (0.75)^{12} = 13 \cdot 0.25 \cdot (0.75)^{12}.$

$P(X = 1) = 13 \cdot 0.25 \cdot (0.75)^{12} \approx 0.05782.$

For $X = 2$:

$P(X = 2) = \binom{13}{2} (0.25)^2 (0.75)^{11} = \frac{13 \cdot 12}{2} \cdot (0.25)^2 \cdot (0.75)^{11}.$

$P(X = 2) = 78 \cdot 0.0625 \cdot (0.75)^{11} \approx 0.14403.$

Step 2: Add probabilities

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

$P(X \leq 2) \approx 0.01336 + 0.05782 + 0.14403 = 0.21521.$

Final Answer:

The probability of rolling a "2" at most 2 times is approximately:

$\boxed{0.2152 \, \text{(rounded to 4 decimal places)}}.$

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Would you like further details or a breakdown of any step? Here are some related questions:

1. How is the binomial distribution used in real-world problems?
2. What does the shape of the binomial probability mass function look like for $n = 13$ and $p = 0.25$?
3. How do we calculate probabilities for $P(X > 2)$ using the complement rule?
4. What is the expected value and standard deviation for $X$ in this distribution?
5. Can we approximate this binomial distribution with a normal distribution?

Tip: When dealing with cumulative probabilities in binomial distributions, consider using statistical tables or software for efficiency.