Q21 / 48 · JEE Advanced 2026 (Paper 1) · Mathematics
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?
Then which of the following statements is (are) TRUE?
- Correct answerAThe events $E_1$ and $F_1$ are independent
- BThe events $E_2$ and $F_2$ are dependent
- Correct answerCThe conditional probability $P(F_1 \mid E_1)$ is equal to the conditional probability $P(F_1 \mid E_2)$
- DThe conditional probability $P(F_1 \mid E_1)$ is greater than the conditional probability $P(F_2 \mid E_2)$
Solution
Total balls in Box I $= 6 + 9 = 15$
Total balls in Box II $= 8 + 12 = 20$
Total balls in the mixture $= 15 + 20 = 35$
Total red balls $= 6 + 8 = 14$
Total green balls $= 9 + 12 = 21$
The probabilities of the given events are:
$P(E_1) = \dfrac{15}{35} = \dfrac{3}{7}$
$P(E_2) = \dfrac{20}{35} = \dfrac{4}{7}$
$P(F_1) = \dfrac{14}{35} = \dfrac{2}{5}$
$P(F_2) = \dfrac{21}{35} = \dfrac{3}{5}$
The joint probabilities are:
$P(E_1 \cap F_1) = \dfrac{6}{35}$
$P(E_2 \cap F_2) = \dfrac{12}{35}$
For statement (A):
$P(E_1) P(F_1) = \dfrac{3}{7} \times \dfrac{2}{5} = \dfrac{6}{35} = P(E_1 \cap F_1)$
Thus, $E_1$ and $F_1$ are independent. Statement (A) is TRUE.
For statement (B):
$P(E_2) P(F_2) = \dfrac{4}{7} \times \dfrac{3}{5} = \dfrac{12}{35} = P(E_2 \cap F_2)$
Thus, $E_2$ and $F_2$ are independent. Statement (B) is FALSE.
For statement (C):
$P(F_1 \mid E_1) = \dfrac{P(E_1 \cap F_1)}{P(E_1)} = \dfrac{6/35}{15/35} = \dfrac{2}{5}$
$P(F_1 \mid E_2) = \dfrac{P(E_2 \cap F_1)}{P(E_2)} = \dfrac{8/35}{20/35} = \dfrac{2}{5}$
Since $P(F_1 \mid E_1) = P(F_1 \mid E_2)$, statement (C) is TRUE.
For statement (D):
$P(F_2 \mid E_2) = \dfrac{P(E_2 \cap F_2)}{P(E_2)} = \dfrac{12/35}{20/35} = \dfrac{3}{5}$
Since $\dfrac{2}{5} < \dfrac{3}{5}$, $P(F_1 \mid E_1)$ is not greater than $P(F_2 \mid E_2)$. Statement (D) is FALSE.
Answer: The events $E_1$ and $F_1$ are independent; The conditional probability $P(F_1 \mid E_1)$ is equal to the conditional probability $P(F_1 \mid E_2)$
Total balls in Box II $= 8 + 12 = 20$
Total balls in the mixture $= 15 + 20 = 35$
Total red balls $= 6 + 8 = 14$
Total green balls $= 9 + 12 = 21$
The probabilities of the given events are:
$P(E_1) = \dfrac{15}{35} = \dfrac{3}{7}$
$P(E_2) = \dfrac{20}{35} = \dfrac{4}{7}$
$P(F_1) = \dfrac{14}{35} = \dfrac{2}{5}$
$P(F_2) = \dfrac{21}{35} = \dfrac{3}{5}$
The joint probabilities are:
$P(E_1 \cap F_1) = \dfrac{6}{35}$
$P(E_2 \cap F_2) = \dfrac{12}{35}$
For statement (A):
$P(E_1) P(F_1) = \dfrac{3}{7} \times \dfrac{2}{5} = \dfrac{6}{35} = P(E_1 \cap F_1)$
Thus, $E_1$ and $F_1$ are independent. Statement (A) is TRUE.
For statement (B):
$P(E_2) P(F_2) = \dfrac{4}{7} \times \dfrac{3}{5} = \dfrac{12}{35} = P(E_2 \cap F_2)$
Thus, $E_2$ and $F_2$ are independent. Statement (B) is FALSE.
For statement (C):
$P(F_1 \mid E_1) = \dfrac{P(E_1 \cap F_1)}{P(E_1)} = \dfrac{6/35}{15/35} = \dfrac{2}{5}$
$P(F_1 \mid E_2) = \dfrac{P(E_2 \cap F_1)}{P(E_2)} = \dfrac{8/35}{20/35} = \dfrac{2}{5}$
Since $P(F_1 \mid E_1) = P(F_1 \mid E_2)$, statement (C) is TRUE.
For statement (D):
$P(F_2 \mid E_2) = \dfrac{P(E_2 \cap F_2)}{P(E_2)} = \dfrac{12/35}{20/35} = \dfrac{3}{5}$
Since $\dfrac{2}{5} < \dfrac{3}{5}$, $P(F_1 \mid E_1)$ is not greater than $P(F_2 \mid E_2)$. Statement (D) is FALSE.
Answer: The events $E_1$ and $F_1$ are independent; The conditional probability $P(F_1 \mid E_1)$ is equal to the conditional probability $P(F_1 \mid E_2)$
Hence, the correct answer is options: 'A', 'C'.