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 E₁ be the event that the ball chosen belonged to Box I and let E₂ be the event that the ball chosen belonged to Box II. Let F₁ be the event that the ball chosen is red and let F₂ be the event that the ball chosen is green. Then which of the following statements is (are) TRUE? [JEE Adv 2026 P1]

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 $P(F_2 | E_2)$

Solution:

Concept:

Independence of events and conditional probability are used to evaluate the given statements.

Explanation:

Total balls: Box I has

15

, Box II has

20

. Combined total is

35

Probabilities are computed as fractions of

35

Check each option:

Option A:

P(E_1) \cdot P(F_1) = \frac{15}{35} \times \frac{14}{35} = \frac{6}{35} = P(E_1 \cap F_1)

. So

E_1

and

F_1

are independent. True.

Option B:

P(E_2) \cdot P(F_2) = \frac{20}{35} \times \frac{21}{35} = \frac{12}{35} = P(E_2 \cap F_2)

. Hence

E_2

and

F_2

are independent, not dependent. False.

Option C:

P(F_1 | E_1) = \frac{6/35}{15/35} = \frac{6}{15} = \frac{2}{5}

P(F_1 | E_2) = \frac{8/35}{20/35} = \frac{8}{20} = \frac{2}{5}

. They are equal. True.

Option D:

P(F_2 | E_2) = \frac{12/35}{20/35} = \frac{12}{20} = \frac{3}{5}

P(F_1 | E_1) = \frac{2}{5}

. Since

\frac{2}{5} < \frac{3}{5}

, the statement is false.

Answer:

Options (A) and (C) are true.

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