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: Find total balls and probabilities.
Box I: \[ 6\ \mathrm{red},\quad 9\ \mathrm{green} \] Total: \[ 15 \] Box II: \[ 8\ \mathrm{red},\quad 12\ \mathrm{green} \] Total: \[ 20 \] Overall total: \[ 35 \]

Step 2: Check Option (A).
\[ P(E_1)=\frac{15}{35}=\frac37 \] Total red balls: \[ 6+8=14 \] Thus: \[ P(F_1)=\frac{14}{35}=\frac25 \] Now: \[ P(E_1\cap F_1)=\frac6{35} \] Also: \[ P(E_1)P(F_1)=\frac37\times\frac25 \] \[ =\frac6{35} \] Since: \[ P(E_1\cap F_1)=P(E_1)P(F_1) \] events are independent. Therefore: \[ \Rightarrow \mathrm{Option\ (A)\ is\ Correct} \]

Step 3: Check Option (B).
\[ P(E_2)=\frac{20}{35}=\frac47 \] Total green balls: \[ 9+12=21 \] \[ P(F_2)=\frac{21}{35}=\frac35 \] Now: \[ P(E_2\cap F_2)=\frac{12}{35} \] And: \[ P(E_2)P(F_2)=\frac47\times\frac35 \] \[ =\frac{12}{35} \] Thus: \[ E_2\ \mathrm{and}\ F_2 \] are independent. Therefore: \[ \Rightarrow \mathrm{Option\ (B)\ is\ Incorrect} \]

Step 4: Check Option (C).
\[ P(F_1|E_1)=\frac6{15}=\frac25 \] \[ P(F_1|E_2)=\frac8{20}=\frac25 \] Hence: \[ P(F_1|E_1)=P(F_1|E_2) \] Therefore: \[ \Rightarrow \mathrm{Option\ (C)\ is\ Correct} \]

Step 5: Check Option (D).
\[ P(F_1|E_1)=\frac25 \] \[ P(F_2|E_2)=\frac{12}{20}=\frac35 \] Since: \[ \frac25<\frac35 \] Therefore: \[ \Rightarrow \mathrm{Option\ (D)\ is\ Incorrect} \]

Step 6: Identify the correct options.
Hence: \[ \boxed{\mathrm{(A)\ and\ (C)}} \]

Answer 2

Concept:

• If every box has the same ratio of red to green balls, then knowing which box a ball came from gives no extra information about its colour -- the two events are automatically independent, with no computation needed.
• Formally, two events A and B are independent when $P(B|A)=P(B)$, meaning the conditional probability equals the plain, unconditional probability.
• Comparing the conditional probability from each box directly is usually faster than checking $P(A \cap B)=P(A)P(B)$ term by term.

Step 1: Write the red-to-green ratio for each box in simplest form.
Box I: $6$ red to $9$ green $=2:3$.
Box II: $8$ red to $12$ green $=2:3$.
Both boxes carry the exact same $2:3$ ratio.

Step 2: Read off what the matching ratio means.
Since red makes up $\frac{2}{5}$ of the balls in both boxes, the conditional probability of red is the same no matter which box is picked. This already signals that colour and box choice do not affect each other.

Step 3: Confirm statement (A) using $P(F_1|E_1)=P(F_1)$.
$P(F_1|E_1)=\frac{6}{15}=\frac{2}{5}$.
Total red balls overall: $6+8=14$ out of $35$, so $P(F_1)=\frac{14}{35}=\frac{2}{5}$.
Since $P(F_1|E_1)=P(F_1)$, $E_1$ and $F_1$ are independent. Statement (A) is TRUE.

Step 4: Confirm statement (B) the same way.
$P(F_2|E_2)=\frac{12}{20}=\frac{3}{5}$.
Total green balls overall: $9+12=21$ out of $35$, so $P(F_2)=\frac{21}{35}=\frac{3}{5}$.
Since $P(F_2|E_2)=P(F_2)$, $E_2$ and $F_2$ are independent, not dependent. Statement (B) is FALSE.

Step 5: Compare $P(F_1|E_1)$ with $P(F_1|E_2)$ for statement (C).
Both boxes share the $2:3$ ratio, so $P(F_1|E_1)=P(F_1|E_2)=\frac{2}{5}$. Statement (C) is TRUE.

Step 6: Compare $P(F_1|E_1)$ with $P(F_2|E_2)$ for statement (D).
$P(F_1|E_1)=\frac{2}{5}$ and $P(F_2|E_2)=\frac{3}{5}$. Since $\frac{2}{5}$ is less than $\frac{3}{5}$, statement (D) is FALSE.

Final Answer: Statements (A) and (C) are the correct ones.