Correct Option: A Step 1: Calculate the total number of ways to draw 3 balls from the 10 balls in a single experiment. Total number of balls = 10 (4 Red + 6 Blue). Number of ways to choose 3 balls from 10 is $\binom{10}{3}$. $\binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 10 \times 3 \times 4 = 120$ ways. Step 2: Calculate the probability of drawing 'no red balls' (i.e., all 3 blue balls) in a single experiment. To draw no red balls, all 3 balls must be blue. Number of ways to choose 3 blue balls from 6 blue balls is $\binom{6}{3}$. $\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$ ways. Probability of drawing no red balls, $P(\text{no R}) = \frac{\text{Number of ways to draw 3 blue balls}}{\text{Total number of ways to draw 3 balls}} = \frac{20}{120} = \frac{1}{6}$. Step 3: Calculate the probability of drawing 'at least one red ball' in a single experiment. $P(\text{at least one R}) = 1 - P(\text{no R})$. $P(\text{at least one R}) = 1 - \frac{1}{6} = \frac{5}{6}$. Let this probability be $p = \frac{5}{6}$. The probability of not drawing at least one red ball (i.e., drawing no red balls) is $q = 1 - p = \frac{1}{6}$. Step 4: Apply the binomial probability formula for multiple independent experiments. The experiment is conducted 3 times independently, and we want the probability that 'at least one red ball is drawn' in exactly two of these three experiments. This is a binomial probability scenario, where $n=3$ (number of trials) and $k=2$ (number of successes). The binomial probability formula is $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$. Substituting the values: $P(X=2) = \binom{3}{2} \left(\frac{5}{6}\right)^2 \left(\frac{1}{6}\right)^{3-2}$ $P(X=2) = 3 \times \left(\frac{25}{36}\right) \times \left(\frac{1}{6}\right)$ $P(X=2) = \frac{3 \times 25}{36 \times 6} = \frac{75}{216}$ Step 5: Simplify the result. Divide the numerator and denominator by their greatest common divisor, which is 3. $\frac{75}{216} = \frac{75 \div 3}{216 \div 3} = \frac{25}{72}$. Thus, the probability that in exactly two of these three experiments, at least one red ball is drawn is $\frac{25}{72}$.