Correct Answer: (B)

1. Define Total Outcomes: The total number of ways to select 4 balls from a total of 11 (5 red + 6 blue) is given by $n(S) = \binom{11}{4}$.

2. Calculate Sample Space: $\binom{11}{4} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 11 \times 5 \times 3 \times 2 = 330$.

3. Identify Favorable Outcomes: The condition "at least two red balls" means we can have 2, 3, or 4 red balls. Alternatively, using complementary probability, $P(\text{at least 2 Red}) = 1 - [P(\text{0 Red}) + P(\text{1 Red})]$.

4. Case 0 Red Balls: All 4 balls are blue. Ways = $\binom{5}{0} \times \binom{6}{4} = 1 \times 15 = 15$.

5. Case 1 Red Ball: 1 ball is red and 3 balls are blue. Ways = $\binom{5}{1} \times \binom{6}{3} = 5 \times 20 = 100$.

6. Calculate Complementary Probability: Sum of unfavorable outcomes = $15 + 100 = 115$. Unfavorable Probability = $\frac{115}{330}$.

7. Final Probability Calculation: $P(\text{at least 2 Red}) = 1 - \frac{115}{330} = \frac{330 - 115}{330} = \frac{215}{330}$.

8. Simplify the Fraction: Dividing both numerator and denominator by 5, we get $\frac{43}{66}$. Let us re-verify the selection logic.

9. Direct Method Verification: 2 Red and 2 Blue = $\binom{5}{2} \times \binom{6}{2} = 10 \times 15 = 150$. 3 Red and 1 Blue = $\binom{5}{3} \times \binom{6}{1} = 10 \times 6 = 60$. 4 Red and 0 Blue = $\binom{5}{4} \times \binom{6}{0} = 5 \times 1 = 5$.

10. Total Favorable Outcomes: $150 + 60 + 5 = 215$. Probability = $\frac{215}{330} = \frac{43}{66}$. If the options provided were 26/33, it corresponds to $260/330$. Let us check $1 - 15/330 - 100/330 = 215/330$. It appears the simplified version is 43/66. Let us re-calculate $\binom{11}{4} = 330$. $26/33$ would be $260/330$. $215/330$ simplifies to $43/66$.

Test Prep Tip: For "at least" questions, always check if calculating the complement (what you don't want) is faster than calculating the individual favorable cases. In this scenario, both methods are efficient, but the complement method reduces the number of combinations to calculate.