Correct Option: C To determine the probability, we first calculate the total number of possible outcomes and then the number of favorable outcomes. **Step 1: Calculate the total number of possible arrangements.** We are selecting 5 distinct balls from 10 distinct balls and then arranging them. This is a permutation problem. The total number of ways to select 5 balls from 10 and arrange them is given by $P(10, 5)$. $P(10, 5) = \frac{10!}{(10-5)!} = \frac{10!}{5!} = 10 \times 9 \times 8 \times 7 \times 6 = 30,240$. **Step 2: Calculate the number of favorable arrangements.** A favorable arrangement is one where all the red balls selected appear consecutively. We need to consider all possible cases for the number of red balls ($r$) and blue balls ($b$) in the selection of 5 balls ($r+b=5$). The selected balls are distinct. Let $N_R=4$ be the number of red balls and $N_B=6$ be the number of blue balls. Case 1: 0 Red balls, 5 Blue balls ($r=0, b=5$) * Number of ways to select: $\binom{4}{0} \times \binom{6}{5} = 1 \times 6 = 6$ ways. * For each such selection (e.g., $B_1, B_2, B_3, B_4, B_5$), all red balls (none) are trivially consecutive. The 5 distinct blue balls can be arranged in $5!$ ways. * Favorable arrangements for this case: $6 \times 5! = 6 \times 120 = 720$. Case 2: 1 Red ball, 4 Blue balls ($r=1, b=4$) * Number of ways to select: $\binom{4}{1} \times \binom{6}{4} = 4 \times 15 = 60$ ways. * For each such selection (e.g., $R_1, B_1, B_2, B_3, B_4$), the single red ball is always consecutive. The 5 distinct selected balls can be arranged in $5!$ ways. * Favorable arrangements for this case: $60 \times 5! = 60 \times 120 = 7,200$. Case 3: 2 Red balls, 3 Blue balls ($r=2, b=3$) * Number of ways to select: $\binom{4}{2} \times \binom{6}{3} = 6 \times 20 = 120$ ways. * For each such selection (e.g., $R_1, R_2, B_1, B_2, B_3$), treat the 2 distinct red balls as a single block ($R_R$). Within this block, the red balls can be arranged in $2!$ ways. Now we arrange this block and the 3 distinct blue balls. This is an arrangement of 4 items ($R_R, B_1, B_2, B_3$), which can be done in $4!$ ways. * Favorable arrangements for each selection: $4! \times 2! = 24 \times 2 = 48$ ways. * Favorable arrangements for this case: $120 \times 48 = 5,760$. Case 4: 3 Red balls, 2 Blue balls ($r=3, b=2$) * Number of ways to select: $\binom{4}{3} \times \binom{6}{2} = 4 \times 15 = 60$ ways. * For each such selection (e.g., $R_1, R_2, R_3, B_1, B_2$), treat the 3 distinct red balls as a single block ($R_R$). Within this block, the red balls can be arranged in $3!$ ways. Now we arrange this block and the 2 distinct blue balls. This is an arrangement of 3 items ($R_R, B_1, B_2$), which can be done in $3!$ ways. * Favorable arrangements for each selection: $3! \times 3! = 6 \times 6 = 36$ ways. * Favorable arrangements for this case: $60 \times 36 = 2,160$. Case 5: 4 Red balls, 1 Blue ball ($r=4, b=1$) * Number of ways to select: $\binom{4}{4} \times \binom{6}{1} = 1 \times 6 = 6$ ways. * For each such selection (e.g., $R_1, R_2, R_3, R_4, B_1$), treat the 4 distinct red balls as a single block ($R_R$). Within this block, the red balls can be arranged in $4!$ ways. Now we arrange this block and the 1 distinct blue ball. This is an arrangement of 2 items ($R_R, B_1$), which can be done in $2!$ ways. * Favorable arrangements for each selection: $2! \times 4! = 2 \times 24 = 48$ ways. * Favorable arrangements for this case: $6 \times 48 = 288$. **Step 3: Sum the favorable arrangements from all cases.** Total favorable arrangements = $720 + 7,200 + 5,760 + 2,160 + 288 = 16,128$. **Step 4: Calculate the probability.** Probability = $\frac{\text{Total Favorable Arrangements}}{\text{Total Possible Arrangements}} = \frac{16,128}{30,240}$. To simplify the fraction: $\frac{16,128}{30,240} = \frac{8,064}{15,120} = \frac{4,032}{7,560} = \frac{2,016}{3,780} = \frac{1,008}{1,890} = \frac{504}{945}$ Divide by 9: $\frac{504 \div 9}{945 \div 9} = \frac{56}{105}$ Divide by 7: $\frac{56 \div 7}{105 \div 7} = \frac{8}{15}$. Thus, the probability is $\frac{8}{15}$. The final answer is $\boxed{\text{C}}$