Question

Suppose that Box I contains \(6\) red balls and \(9\) green balls, and Box II contains \(8\) red balls and \(12\) green balls. All the balls of Box I and Box II are mixed together and a ball is chosen at random from them. Let \(E_1\) be the event that the ball chosen belonged to Box I and let \(E_2\) be the event that the ball chosen belonged to Box II. Let \(F_1\) be the event that the ball chosen is red and let \(F_2\) be the event that the ball chosen is green. Then which of the following statements is (are) TRUE?

• A. The events \(E_1\) and \(F_1\) are independent
• B. The events \(E_2\) and \(F_2\) are dependent
• C. The conditional probability \(P(F_1|E_1)\) is equal to the conditional probability \(P(F_1|E_2)\)
• D. The conditional probability \(P(F_1|E_1)\) is greater than the conditional probability \(P(F_2|E_2)\)

Hint:

Before doing any big calculation, check whether every box has the same red-to-green ratio. If it does, picking a box tells you nothing about colour, and that one observation answers most of the independence-type statements immediately.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Two events \( A \) and \( B \) are independent if \( P(A|B) = P(A) \). In the context of probability experiments with subsets (boxes), independence occurs if the proportion of the desired trait (e.g., color) is exactly the same across all subsets.
Step 2: Key Formula or Approach:
Box I: Red (\( R \)) = 6, Green (\( G \)) = 9, Total (\( T_1 \)) = 15.
Box II: \( R \) = 8, \( G \) = 12, Total (\( T_2 \)) = 20.
Total balls mixed = 35.
Step 3: Detailed Explanation:
1. Probabilities of Box Selection:
Since there are 15 balls in Box I and 20 in Box II, and all are mixed:
\( P(E_1) = 15/35 = 3/7 \).
\( P(E_2) = 20/35 = 4/7 \).
2. Conditional Probabilities of Colors:
If we know a ball is from Box I: \( P(F_1 | E_1) = 6/15 = 2/5 = 0.4 \).
If we know a ball is from Box II: \( P(F_1 | E_2) = 8/20 = 2/5 = 0.4 \).
Since \( P(F_1 | E_1) = P(F_1 | E_2) \), statement (C) is True.
3. Marginal Probabilities:
\( P(F_1) = P(E_1)P(F_1|E_1) + P(E_2)P(F_1|E_2) = (3/7)(2/5) + (4/7)(2/5) = 2/5 \).
\( P(F_2) = 1 - 2/5 = 3/5 \).
4. Independence check (A and B):
\( P(F_1 | E_1) = 2/5 = P(F_1) \). This implies \( E_1 \) and \( F_1 \) are independent. Statement (A) is True.
Similarly, \( P(F_2 | E_2) = 12/20 = 3/5 = P(F_2) \). This implies \( E_2 \) and \( F_2 \) are independent. Statement (B) is False.
5. Reverse Conditional check (D):
\( P(E_1 | F_1) = \frac{P(E_1 \cap F_1)}{P(F_1)} = \frac{(3/7)(2/5)}{2/5} = 3/7 \approx 0.428 \).
\( P(E_2 | F_2) = \frac{P(E_2 \cap F_2)}{P(F_2)} = \frac{(4/7)(3/5)}{3/5} = 4/7 \approx 0.571 \).
\( 3/7<4/7 \), so statement (D) is False.
Step 4: Final Answer:
Statements (A) and (C) are the correct choices.