To calculate the probability, we can use the binomial probability formula:

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

Where:

• $n$ = total number of trials (4 dice rolled)
• $k$ = number of successes (3 dice landing on a 3)
• $p$ = probability of success in a single trial ($\frac{1}{6}$ for a 3)
• $\binom{n}{k}$ = combination formula, $\frac{n!}{k!(n-k)!}$

Step 1: Plugging in the values

• $n = 4$
• $k = 3$
• $p = \frac{1}{6}$

$P(X = 3) = \binom{4}{3} \left(\frac{1}{6}\right)^3 \left(\frac{5}{6}\right)^1$

Step 2: Calculate the combination

$\binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4}{1} = 4$

Step 3: Calculate the probabilities

$P(X = 3) = 4 \cdot \left(\frac{1}{6}\right)^3 \cdot \left(\frac{5}{6}\right)$

$P(X = 3) = 4 \cdot \frac{1}{216} \cdot \frac{5}{6}$

$P(X = 3) = 4 \cdot \frac{5}{1296}$

$P(X = 3) = \frac{20}{1296}$

Step 4: Simplify the fraction

$P(X = 3) = \frac{5}{324} \approx 0.0154$

Step 5: Match the closest answer

The closest probability is 0.02.

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Do you want details or have any questions? Here are five related questions for deeper understanding:

1. How is the binomial probability formula derived?
2. What is the probability of exactly 2 cubes landing on a 3?
3. How would the calculation change if the dice were biased?
4. What is the probability that none of the dice lands on a 3?
5. How does the combination factor affect the result in binomial probabilities?

Tip: The binomial formula is great for solving problems where you count specific outcomes in a series of independent trials!