JEE Advanced 2026 — Paper 1 — MATH-II solution

JEE Advanced2026Paper 1MATH-II

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 $P (F_{2} ∣ E_{2})$

Solution

Pooled counts: Box I has 15 balls (6 red, 9 green), Box II has 20 balls (8 red, 12 green). Total $= 35$ balls, $14$ red, $21$ green.

Marginals: $P (E_{1}) = \frac{15}{35} = \frac{3}{7}$ , $P (E_{2}) = \frac{4}{7}$ , $P (F_{1}) = \frac{14}{35} = \frac{2}{5}$ , $P (F_{2}) = \frac{3}{5}$ .

Conditionals:

$P (F_{1} ∣ E_{1}) = \frac{6}{15} = \frac{2}{5}$ , $P (F_{1} ∣ E_{2}) = \frac{8}{20} = \frac{2}{5}$

$P (F_{2} ∣ E_{2}) = \frac{12}{20} = \frac{3}{5}$

(A) Since $P (F_{1} ∣ E_{1}) = \frac{2}{5} = P (F_{1})$ , the events $E_{1}$ and $F_{1}$ are independent. TRUE.

(B) Since $P (F_{2} ∣ E_{2}) = \frac{3}{5} = P (F_{2})$ , $E_{2}$ and $F_{2}$ are independent, not dependent. FALSE.

(C) Both equal $\frac{2}{5}$ . TRUE.

(D) $P (F_{1} ∣ E_{1}) = \frac{2}{5} < \frac{3}{5} = P (F_{2} ∣ E_{2})$ . FALSE.

$∴$ The correct options are (A) and (C).

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