1. **State the problem:** We roll a six-sided die 12 times and want to find the probability of getting the number 4 exactly 5 times. 2. **Identify the distribution:** This is a binomial probability problem where each roll is a trial with two outcomes: getting a 4 (success) or not getting a 4 (failure). 3. **Formula:** The binomial probability formula is $$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$$ where: - $n = 12$ (number of trials) - $k = 5$ (number of successes) - $p = \frac{1}{6}$ (probability of success on each trial, since one face out of six is a 4) 4. **Calculate the binomial coefficient:** $$\binom{12}{5} = \frac{12!}{5! \times (12-5)!} = \frac{12!}{5! \times 7!} = 792$$ 5. **Calculate the probability:** $$P(X=5) = 792 \times \left(\frac{1}{6}\right)^5 \times \left(\frac{5}{6}\right)^7$$ 6. **Simplify:** - $\left(\frac{1}{6}\right)^5 = \frac{1}{7776}$ - $\left(\frac{5}{6}\right)^7 = \frac{78125}{279936}$ So, $$P(X=5) = 792 \times \frac{1}{7776} \times \frac{78125}{279936}$$ 7. **Final calculation:** $$P(X=5) = \frac{792 \times 78125}{7776 \times 279936} \approx 0.0283$$ **Answer:** The probability of rolling a 4 exactly 5 times in 12 rolls is approximately **0.0283** or 2.83%.