To compute the probability of rolling exactly four 2's in 10 rolls of a standard 4-sided die, we'll use the binomial probability formula.

Steps:

1. Identify the parameters:

   • Number of trials ($n$): 10
   • Number of successes ($k$): 4 (rolling a 2 is considered a "success")
   • Probability of success on a single trial ($p$): $\frac{1}{4} = 0.25$
   • Probability of failure on a single trial ($1 - p$): $\frac{3}{4} = 0.75$

2. Use the binomial probability formula:

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

where: $\binom{n}{k} = \frac{n!}{k!(n-k)!}$

3. Substitute the values:

$P(X = 4) = \binom{10}{4} \times (0.25)^4 \times (0.75)^6$

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

4. Calculate the probability:

$P(X = 4) = 210 \times (0.25)^4 \times (0.75)^6$

$P(X = 4) = 210 \times (0.00390625) \times (0.177978515625)$

$P(X = 4) = 210 \times 0.00069427490234375$

$P(X = 4) \approx 0.146$

So, the probability of rolling exactly four 2's in 10 rolls of a 4-sided die is 0.146 (rounded to three decimal places).

Would you like any more details or have any questions?

Here are five related questions:

1. What is the probability of getting exactly five 2's in 10 rolls of a 4-sided die?
2. How would the probability change if you wanted to roll exactly three 2's instead of four?
3. What is the probability of getting no 2's 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 1's in 10 rolls of a 4-sided die?

Tip: The binomial distribution is very useful when dealing with independent trials, like rolling a die multiple times, where each outcome is either a "success" or "failure".