Correct Answer: (D) Step-by-step Solution To find the probability that at least two of the four balls drawn are red, we can calculate the total number of possible outcomes and then find the sum of the favorable outcomes (where 2, 3, or 4 balls are red). 1. Total Number of Outcomes The total number of balls in the box is $5 \text{ (red)} + 6 \text{ (blue)} = 11$. Since we are drawing 4 balls at random, the total number of ways to choose them is:$$\binom{11}{4} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330$$ 2. Number of Favorable Outcomes We want "at least two red balls," which means we can have exactly 2, 3, or 4 red balls. Case 1: Exactly 2 Red Balls (and 2 Blue)$$\binom{5}{2} \times \binom{6}{2} = 10 \times 15 = 150$$ Case 2: Exactly 3 Red Balls (and 1 Blue)$$\binom{5}{3} \times \binom{6}{1} = 10 \times 6 = 60$$ Case 3: Exactly 4 Red Balls (and 0 Blue)$$\binom{5}{4} \times \binom{6}{0} = 5 \times 1 = 5$$Total favorable outcomes $= 150 + 60 + 5 = 215$. 3. Final Probability The probability is the ratio of favorable outcomes to total outcomes:$$P(\text{at least 2 red}) = \frac{215}{330}$$ Simplifying the fraction by dividing both numerator and denominator by 5:$$P(\text{at least 2 red}) = \frac{43}{66}$$ The probability that at least two of the balls are red is $\frac{43}{66}$.