1. Key Concepts and Formulas

This problem is a quintessential application of conditional probability, specifically utilizing Bayes' Theorem and the Law of Total Probability. These concepts are fundamental for analyzing situations where we need to update probabilities based on new information.

• Bayes' Theorem: This theorem helps us calculate the probability of a "cause" given an "effect." If $A_1, A_2, \dots, A_n$ are mutually exclusive and exhaustive events (potential causes) and $B$ is an observed event (effect), then the probability of a specific cause $A_k$ given the effect $B$ is: $P(A_k|B) = \frac{P(B|A_k) \cdot P(A_k)}{P(B)}$
• Law of Total Probability: This law helps us find the overall probability of an event $B$ by summing the probabilities of $B$ occurring under each of the mutually exclusive and exhaustive conditions $A_1, A_2, \dots, A_n$: $P(B) = P(B|A_1)P(A_1) + P(B|A_2)P(A_2) + \dots + P(B|A_n)P(A_n)$
• Conditional Probability (Basic): The probability of an event $E_1$ occurring given that event $E_2$ has already occurred is $P(E_1|E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)}$. In scenarios involving drawing from bags, $P(\text{color}|\text{bag})$ is the ratio of balls of that color in the bag to the total number of balls in that bag.

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2. Step-by-Step Solution

To solve this problem systematically, we will first define all relevant events and their initial probabilities, then apply the aforementioned theorems to calculate the required conditional probabilities $p$ and $q$.

Step 1: Define Events and Initial Probabilities

The first crucial step is to clearly define the events involved in the problem. This makes the application of probability formulas straightforward and reduces confusion.

Let's define the following events:

• $B_1$: The event that Bag I is chosen.
• $B_2$: The event that Bag II is chosen.
• $B_3$: The event that Bag III is chosen.
• $R$: The event that a Red ball is drawn.
• $G$: The event that a Green ball is drawn.

The problem states that a person chooses a bag at random from three identical bags. This implies that each bag has an equal chance of being selected. Therefore, the prior probabilities of choosing each bag are: $P(B_1) = P(B_2) = P(B_3) = \frac{1}{3}$

Next, we extract the conditional probabilities of drawing a specific color ball from each bag, based on the given table. Each bag contains a total of 10 balls.

• From Bag I (3 Red, 2 Blue, 5 Green):

  • Probability of drawing a Red ball given Bag I: $P(R|B_1) = \frac{\text{Number of Red balls in Bag I}}{\text{Total balls in Bag I}} = \frac{3}{10}$
  • Probability of drawing a Green ball given Bag I: $P(G|B_1) = \frac{\text{Number of Green balls in Bag I}}{\text{Total balls in Bag I}} = \frac{5}{10}$

• From Bag II (4 Red, 3 Blue, 3 Green):

  • Probability of drawing a Red ball given Bag II: $P(R|B_2) = \frac{4}{10}$
  • Probability of drawing a Green ball given Bag II: $P(G|B_2) = \frac{3}{10}$

• From Bag III (5 Red, 1 Blue, 4 Green):

  • Probability of drawing a Red ball given Bag III: $P(R|B_3) = \frac{5}{10}$
  • Probability of drawing a Green ball given Bag III: $P(G|B_3) = \frac{4}{10}$

Step 2: Calculate $p$, the probability that the Red ball is from Bag I

The problem defines $p$ as the probability that the ball is from Bag I, given that the ball drawn is Red. In probability notation, this is $P(B_1 | R)$. This is a direct application of Bayes' Theorem.

2a. Calculate the overall probability of drawing a Red ball, $P(R)$, using the Law of Total Probability. The event of drawing a Red ball ($R$) can happen if we choose Bag I and draw a Red ball, OR choose Bag II and draw a Red ball, OR choose Bag III and draw a Red ball. These are mutually exclusive scenarios. $P(R) = P(R|B_1)P(B_1) + P(R|B_2)P(B_2) + P(R|B_3)P(B_3)$ Substituting the values defined in Step 1: $P(R) = \left(\frac{3}{10} \times \frac{1}{3}\right) + \left(\frac{4}{10} \times \frac{1}{3}\right) + \left(\frac{5}{10} \times \frac{1}{3}\right)$ We can factor out the common term $P(B_i) = \frac{1}{3}$: $P(R) = \frac{1}{3} \left(\frac{3}{10} + \frac{4}{10} + \frac{5}{10}\right)$ $P(R) = \frac{1}{3} \left(\frac{3+4+5}{10}\right) = \frac{1}{3} \left(\frac{12}{10}\right)$ Simplifying the fractions: $P(R) = \frac{1}{3} \times \frac{6}{5} = \frac{6}{15} = \frac{2}{5}$ So, the overall probability of drawing a Red ball is $P(R) = \frac{2}{5}$.

2b. Apply Bayes' Theorem to find $p = P(B_1 | R)$. Now we use Bayes' Theorem with $A_k = B_1$ and $B = R$: $p = P(B_1 | R) = \frac{P(R | B_1) \cdot P(B_1)}{P(R)}$ Substitute the values: $P(R|B_1) = \frac{3}{10}$, $P(B_1) = \frac{1}{3}$, and $P(R) = \frac{2}{5}$. $p = \frac{\left(\frac{3}{10}\right) \cdot \left(\frac{1}{3}\right)}{\frac{2}{5}}$ First, calculate the numerator: $\frac{3}{10} \times \frac{1}{3} = \frac{3}{30} = \frac{1}{10}$. Then, substitute back: $p = \frac{\frac{1}{10}}{\frac{2}{5}} = \frac{1}{10} \times \frac{5}{2} = \frac{5}{20} = \frac{1}{4}$ Thus, $p = \frac{1}{4}$.

Step 3: Calculate $q$, the probability that the Green ball is from Bag III

The problem defines $q$ as the probability that the ball is from Bag III, given that the ball drawn is Green. In probability notation, this is $P(B_3 | G)$. This is another application of Bayes' Theorem.

3a. Calculate the overall probability of drawing a Green ball, $P(G)$, using the Law of Total Probability. Similar to calculating $P(R)$, the event of drawing a Green ball ($G$) can happen if we choose Bag I and draw a Green ball, OR choose Bag II and draw a Green ball, OR choose Bag III and draw a Green ball. $P(G) = P(G|B_1)P(B_1) + P(G|B_2)P(B_2) + P(G|B_3)P(B_3)$ Substituting the values defined in Step 1: $P(G) = \left(\frac{5}{10} \times \frac{1}{3}\right) + \left(\frac{3}{10} \times \frac{1}{3}\right) + \left(\frac{4}{10} \times \frac{1}{3}\right)$ Factor out the common term $P(B_i) = \frac{1}{3}$: $P(G) = \frac{1}{3} \left(\frac{5}{10} + \frac{3}{10} + \frac{4}{10}\right)$ $P(G) = \frac{1}{3} \left(\frac{5+3+4}{10}\right) = \frac{1}{3} \left(\frac{12}{10}\right)$ Simplifying the fractions: $P(G) = \frac{1}{3} \times \frac{6}{5} = \frac{6}{15} = \frac{2}{5}$ So, the overall probability of drawing a Green ball is $P(G) = \frac{2}{5}$. (Self-check: Notice that $P(R) = P(G) = \frac{2}{5}$. This happens because the total number of Red balls across all bags (12) is equal to the total number of Green balls across all bags (12), and the bag selection probability is uniform.)

3b. Apply Bayes' Theorem to find $q = P(B_3 | G)$. Now we use Bayes' Theorem with $A_k = B_3$ and $B = G$: $q = P(B_3 | G) = \frac{P(G | B_3) \cdot P(B_3)}{P(G)}$ Substitute the values: $P(G|B_3) = \frac{4}{10}$, $P(B_3) = \frac{1}{3}$, and $P(G) = \frac{2}{5}$. $q = \frac{\left(\frac{4}{10}\right) \cdot \left(\frac{1}{3}\right)}{\frac{2}{5}}$ First, calculate the numerator: $\frac{4}{10} \times \frac{1}{3} = \frac{4}{30} = \frac{2}{15}$. Then, substitute back: $q = \frac{\frac{2}{15}}{\frac{2}{5}} = \frac{2}{15} \times \frac{5}{2} = \frac{10}{30} = \frac{1}{3}$ Thus, $q = \frac{1}{3}$.

Step 4: Calculate the Final Expression $\left(\frac{1}{p} + \frac{1}{q}\right)$

We have found the values for $p$ and $q$:

• $p = \frac{1}{4}$
• $q = \frac{1}{3}$

Now, we need to calculate $\left(\frac{1}{p} + \frac{1}{q}\right)$: First, find the reciprocals: $\frac{1}{p} = \frac{1}{\frac{1}{4}} = 4$ $\frac{1}{q} = \frac{1}{\frac{1}{3}} = 3$ Finally, add these values: $\left(\frac{1}{p} + \frac{1}{q}\right) = 4 + 3 = 7$

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3. Common Mistakes & Tips

• Confusing Conditional Probabilities: A common error is mixing up $P(A|B)$ with $P(B|A)$. Bayes' Theorem is specifically designed to "reverse" conditional probabilities. Always define your events clearly to avoid this.
• Forgetting Law of Total Probability: The denominator in Bayes' Theorem, $P(B)$, is often not given directly and must be calculated using the Law of Total Probability by summing up all possible ways event $B$ could occur.
• Calculation Errors: Working with fractions requires careful arithmetic. Double-check your additions, multiplications, and divisions, especially when dealing with compound fractions.
• Tip for Setup: Always start by listing all given probabilities and the probabilities you need to find. This structured approach helps in identifying which formula to use at each step.
• Tip for Simplification: Look for common factors to simplify calculations (e.g., $P(B_i) = 1/3$ in this problem). This can save time and reduce errors.

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4. Summary

This problem effectively demonstrates the power of Bayes' Theorem in conjunction with the Law of Total Probability to solve real-world conditional probability scenarios. By systematically defining events, calculating prior probabilities and conditional probabilities from the given data, and then applying these two fundamental theorems, we were able to determine the probabilities $p$ and $q$. Finally, a simple arithmetic calculation yielded the required expression. The structured approach ensures accuracy and clarity in complex probability problems.

The final answer is $\boxed{7}$, which corresponds to option (A).