To solve this problem, we need to compute the probability of rolling exactly four 2s when rolling a standard 4-sided die 10 times.

The number of ways to choose exactly four 2s in 10 rolls is given by the binomial coefficient, denoted as $\binom{n}{k}$, where $n$ is the total number of rolls (10) and $k$ is the number of times you want to roll a 2 (4).

The probability of rolling a 2 on a single roll of a 4-sided die is $\frac{1}{4}$, and the probability of not rolling a 2 is $\frac{3}{4}$.

The probability of exactly four 2s in 10 rolls is given by the binomial distribution formula:

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

where:

• $n = 10$ (number of trials),
• $k = 4$ (number of successful outcomes),
• $p = \frac{1}{4}$ (probability of rolling a 2),
• $1-p = \frac{3}{4}$ (probability of not rolling a 2).

Let's compute this step by step.

$\binom{10}{4} = \frac{10!}{4!(10-4)!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$

$P(X = 4) = 210 \times \left(\frac{1}{4}\right)^4 \times \left(\frac{3}{4}\right)^{6}$

$P(X = 4) = 210 \times \left(\frac{1}{256}\right) \times \left(\frac{729}{4096}\right)$

$P(X = 4) = 210 \times \frac{729}{1048576}$

$P(X = 4) \approx 0.146$

Thus, the probability of getting exactly four 2s in 10 rolls of a standard 4-sided die is approximately 0.146.

Would you like further details on this calculation, or do you have any other questions?

Here are five related questions:

1. What is the probability of getting exactly three 2s in 10 rolls of a 4-sided die?
2. How would the probability change if you wanted to roll exactly four 3s instead?
3. What is the probability of getting at least one 2 in 10 rolls?
4. How does the probability change if you roll the die 12 times instead of 10 times?
5. What is the probability of getting exactly four 2s in 8 rolls of a 4-sided die?

Tip: When working with binomial probabilities, it's useful to familiarize yourself with the binomial distribution and the use of combinations (binomial coefficients).