1. **Problem 4.1:** Two fair dice are rolled. Find the probability that the sum of the scores is a prime number. 2. Each die has outcomes from 1 to 6, so there are $6 \times 6 = 36$ total equally likely outcomes. 3. The possible sums range from 2 to 12. Prime sums in this range are 2, 3, 5, 7, and 11. 4. Count the number of outcomes giving prime sums: - Sum 2: (1,1) — 1 outcome - Sum 3: (1,2),(2,1) — 2 outcomes - Sum 5: (1,4),(2,3),(3,2),(4,1) — 4 outcomes - Sum 7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) — 6 outcomes - Sum 11: (5,6),(6,5) — 2 outcomes 5. Total favorable outcomes = $1+2+4+6+2=15$. 6. Probability sum is prime = $\frac{15}{36} = \frac{5}{12}$. --- 7. **Problem 4.2:** Kia and Rae each attempt to hit a target. $P(Kia\ hit) = \frac{4}{5}$, $P(Rae\ hit) = \frac{3}{4}$. 8. We want probability only one of them hits the target (i.e., exactly one hits). 9. Using independence: - $P(only\ Kia\ hits) = P(Kia\ hits) \times P(Rae\ misses) = \frac{4}{5} \times (1 - \frac{3}{4}) = \frac{4}{5} \times \frac{1}{4} = \frac{1}{5}$. - $P(only\ Rae\ hits) = P(Rae\ hits) \times P(Kia\ misses) = \frac{3}{4} \times (1 - \frac{4}{5}) = \frac{3}{4} \times \frac{1}{5} = \frac{3}{20}$. 10. Total probability only one hits = $\frac{1}{5} + \frac{3}{20} = \frac{4}{20} + \frac{3}{20} = \frac{7}{20}$. --- 11. **Problem 5.1:** A biased die is thrown 30 times, eight sixes observed. Assuming same bias, the probability of six is estimated as $p=\frac{8}{30} = \frac{4}{15}$. 12. The die is thrown another 12 times. Find probability exactly 2 sixes occur. 13. Use binomial distribution: $P(X=k)=\binom{n}{k} p^k (1-p)^{n-k}$. 14. Here $n=12$, $k=2$, $p=\frac{4}{15}$. $$P(X=2) = \binom{12}{2} \left(\frac{4}{15}\right)^2 \left(1-\frac{4}{15}\right)^{10} = 66 \times \left(\frac{16}{225}\right) \times \left(\frac{11}{15}\right)^{10}.$$ 15. This is the exact probability formula. --- 16. **Problem 5.2:** Expected number of sixes in 12 throws is $E(X) = n p = 12 \times \frac{4}{15} = \frac{48}{15} = 3.2$. --- 17. **Problem 5.3:** Variance of number of sixes is $Var(X) = n p (1-p) = 12 \times \frac{4}{15} \times \left(1-\frac{4}{15}\right) = 12 \times \frac{4}{15} \times \frac{11}{15} = \frac{528}{225} = 2.3467$ approximately. **Final answers:** - 4.1 Probability sum prime = $\frac{5}{12}$. - 4.2 Probability only one hits = $\frac{7}{20}$. - 5.1 Probability exactly two sixes in 12 throws = $66 \times \left(\frac{4}{15}\right)^2 \times \left(\frac{11}{15}\right)^{10}$. - 5.2 Expected sixes in 12 throws = $3.2$. - 5.3 Variance in 12 throws = $\frac{528}{225}$ or approx. $2.3467$.