A bag contains some balls of three different colours. Probability of drawing a blue colour ball is $\left(\frac{5}{24}\right)$ and the ratio of red to green colour balls is $9 : 10$. What is the total number of balls in the bag? I. Average of total number of balls of all the three colours is 8. II. Two balls are drawn randomly from the bag and probability that both the balls are green is $\frac{15}{92}$. III. One ball is drawn at random from the bag and the probability that the ball is either blue or red is $\left(\frac{7}{12}\right)$.

### Concept & Probability Proportions To find the total number of balls, we need absolute values since the question only provides ratios and probabilities (which are relative). Let total balls be $T$. $$P(\text{Blue}) = \frac{5}{24} \implies \text{Blue balls} = \frac{5}{24}T$$ $$ \text{Remaining (Red + Green)} = T - \frac{5}{24}T = \frac{19}{24}T $$ Since Red : Green = $9 : 10$, Green balls = $\frac{10}{19} \times \frac{19}{24}T = \frac{10}{24}T$. ### Step-by-Step Solution * **Statement I:** Average of all three colors is 8. $\frac{T}{3} = 8 \implies T = 24$. This gives a unique answer. (Statement I alone is sufficient). * **Statement II:** Probability of drawing two green balls is $\frac{15}{92}$. $$ \frac{\binom{G}{2}}{\binom{T}{2}} = \frac{15}{92} $$ Substitute $G = \frac{10T}{24}$: $$ \frac{\frac{10T}{24}(\frac{10T}{24} - 1)}{T(T - 1)} = \frac{15}{92} $$ Solving this quadratic equation for $T$ yields a single positive integer $T = 24$. (Statement II alone is sufficient). * **Statement III:** Probability of blue or red is $\frac{7}{12}$. From our initial proportions, Red = $\frac{9}{24}T$. $P(\text{Blue or Red}) = \frac{5}{24} + \frac{9}{24} = \frac{14}{24} = \frac{7}{12}$. This is universally true based on the question stem and provides no new information about the total number of balls. (Statement III is not sufficient). ### Exam Strategy & Shortcut In Data Sufficiency, do not solve complex equations unless required to prove a unique solution exists. For Statement II, recognizing that a specific integer probability for dependent events almost always constraints the total pool to a single integer is key to saving time. Notice Statement III just repeats deduced relative data. ### Common Pitfall Mistaking Statement III as new information. Always check if the statement can be mathematically derived from the question stem alone. ### Final Answer Therefore, the correct answer is **Either I alone or II alone is sufficient.**