Directions: Read the following passage and answer the given questions. There are five pouches (A, B, C, D, and E) which contain candies of four flavors (Orange, Pineapple, Caramel, and Coffee). Below table shows the number of candies of four different flavors in five pouches: | | Orange candy | Pineapple candy | Caramel candy | Coffee candy | |---|---|---|---|---| | Pouch A | 12 | --- | --- | --- | | Pouch B | --- | 24 | --- | 10 | | Pouch C | --- | --- | 16 | --- | | Pouch D | --- | --- | 10 | --- | | Pouch E | --- | 10 | --- | --- | Some information is given below: - When one candy is picked at random from pouch A, then the probability of getting one coffee candy is $1/3$ and when one candy is picked at random from pouch C, then the probability of getting neither caramel nor coffee is $1/2$. - In pouch E, number of orange candies is same as number of caramel candies. Total number of caramel candies in all five pouches together is $69$. - The number of pineapple candies in pouches C and D are equal and when two candies are picked at random from pouch A, then the probability of getting both being pineapple candies is $1/22$. - When one candy is picked at random from pouch B, then the probability of getting either orange candy or caramel candy is $25/42$. Total number of caramel candies and coffee candies are same in all five pouches together. Number of caramel candies in pouch D is $2$ more than that in pouch A. - When one candy is picked at random from pouch C, then the probability that the candy picked is not caramel flavor is $4/5$. When two candies are picked at random from pouch D, then the probability that none of the candies is orange flavor is $17/35$. - When one candy is picked at random from pouch E, then the probability of getting one orange candy is $1/6$. When one candy is picked at random from pouch D, then probability of getting coffee candy is $1/5$. When one candy is picked at random from pouch E, then the probability of getting one pineapple candy is $1/3$. The question below contains a statement followed by Quantity I, Quantity II and Quantity III. Find all quantities to find the relationship among them. If 5 orange candies from pouch A are shifted to pouch C and 2 caramel candies from pouch D are shifted to pouch A. **Quantity I:** When two candies are picked at random from pouch D, then what is the probability that the both candies are caramel flavor? **Quantity II:** When three candies are picked at random from pouch A, then what is the probability that exactly two candies are pineapple flavor? **Quantity III:** When one candy is picked at random from pouch C, then what is the probability that the picked candy is orange flavor?

### Concept & Probability Logic To solve this Data Interpretation problem, we first determine the missing values in the table by forming equations from the given probabilities. $$P(\text{Event}) = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}}$$ ### Step-by-Step Solution **Step 1: Complete the Candy Distribution Table** * **Pouch E:** $P(\text{pineapple}) = 1/3 \implies 10/T_E = 1/3 \implies T_E = 30$. Since $P(\text{orange}) = 1/6$, $O_E = 5$. Given $O_E = Ca_E$, $Ca_E = 5$. $Co_E = 30-20 = 10$. * **Pouch D:** $Ca_D = 10$. $P(\text{coffee}) = 1/5 \implies Co_D/T_D = 1/5$. Testing multiples of 5 with $P(\text{none is orange}) = 17/35$ yields $T_D = 50$, $O_D = 15$. Thus, $P_D = 15$, $Co_D = 10$. * **Pouch C:** $Ca_C = 16$. $P(\text{not caramel}) = 4/5 \implies T_C = 80$. $P(\text{neither Ca nor Co}) = 1/2 \implies O_C + P_C = 40$. Since $P_C = P_D = 15$, $O_C = 25$. $Co_C = 24$. * **Pouch A:** $Ca_D = Ca_A + 2 \implies Ca_A = 8$. $P(\text{coffee}) = 1/3 \implies Co_A/T_A = 1/3$. Using $P(\text{two pineapple}) = 1/22$ gives $T_A = 45$, $P_A = 10$, $Co_A = 15$. * **Pouch B:** Total Caramel = 69 $\implies Ca_B = 30$. $P(\text{orange or caramel}) = 25/42 \implies T_B = 84$, leaving $O_B = 20$. **Step 2: Evaluate New Quantities** * **Shift operations:** 5 Orange moved from A to C. 2 Caramel moved from D to A. * **New Totals:** - Pouch A: $T_A = 42$ (Orange=7, Pineapple=10, Caramel=10, Coffee=15) - Pouch C: $T_C = 85$ (Orange=30, Pineapple=15, Caramel=16, Coffee=24) - Pouch D: $T_D = 48$ (Orange=15, Pineapple=15, Caramel=8, Coffee=10) **Step 3: Calculate Given Probabilities** * **Quantity I:** Pick 2 candies from D. $P(\text{both Caramel}) = \frac{^8C_2}{^{48}C_2} = \frac{28}{1128} = \frac{7}{282} \approx 0.024$ * **Quantity II:** Pick 3 candies from A. $P(\text{exactly 2 Pineapple}) = \frac{^{10}C_2 \times ^{32}C_1}{^{42}C_3} = \frac{45 \times 32}{11480} = \frac{36}{287} \approx 0.125$ * **Quantity III:** Pick 1 candy from C. $P(\text{Orange}) = \frac{30}{85} = \frac{6}{17} \approx 0.353$ Comparing the values, Quantity I < Quantity II < Quantity III. ### Exam Strategy & Shortcut Whenever tackling quantity comparison questions with probability, it is not always necessary to solve to the last decimal point. Since $7/282$ is very small, $36/287$ is around $1/8$, and $6/17$ is over $1/3$, the inequality is distinct and can be quickly judged visually once the fractions are simplified. ### Common Pitfall Forgetting to adjust the total number of candies in each pouch after shifting specific flavors between them, causing incorrect denominator calculations. ### Final Answer Therefore, the correct answer is **Quantity I < Quantity II < Quantity III**.