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 $1/3$. In pouch B, 25% of candies are rotten in which one-third is caramel flavor. If two candies are picked at random from pouch B, then what is the probability of getting one rotten caramel candy and another good candy?

### 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: Calculate Rotten and Good Candies in Pouch B** * Total candies in Pouch B = 84. * Rotten candies = 25% of 84 = 21. * Good candies = 84 - 21 = 63. * Rotten Caramel candies = 1/3 of 21 = 7. **Step 3: Find the Required Probability** * We need 1 rotten caramel candy and 1 good candy. * Ways to choose 1 rotten caramel = $^7C_1 = 7$. * Ways to choose 1 good candy = $^{63}C_1 = 63$. * Total ways to choose 2 candies = $^{84}C_2 = \frac{84 \times 83}{2} = 3486$. * Probability = $\frac{7 \times 63}{3486} = \frac{441}{3486} = \frac{21}{166}$. ### Exam Strategy & Shortcut Cancel out common factors before multiplying fully. Notice that $3486 = 42 \times 83$. So, $\frac{7 \times 63}{42 \times 83} = \frac{63}{6 \times 83} = \frac{21}{2 \times 83} = \frac{21}{166}$. Keeping terms factored simplifies the math greatly. ### Common Pitfall A common error is confusing the total pool of candies to select from; ensure you clearly separate "rotten caramel" and the overall "good" candies, ignoring the other flavors of rotten candies. ### Final Answer Therefore, the correct answer is **21/166**.