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?
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?
Show Hint
Two events \(A\) and \(B\) are independent if:
\[
P(A\cap B)=P(A)P(B)
\]
Updated On: May 20, 2026
- The events \(E_1\) and \(F_1\) are independent
- The events \(E_2\) and \(F_2\) are dependent
- The conditional probability \(P(F_1|E_1)\) is equal to the conditional probability \(P(F_1|E_2)\)
- The conditional probability \(P(F_1|E_1)\) is greater than the conditional probability \(P(F_2|E_2)\)
Show Solution
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The Correct Option is A
Solution and Explanation
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: \[ \boxed{\mathrm{(A)\ and\ (C)}} \]
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: \[ \boxed{\mathrm{(A)\ and\ (C)}} \]
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