Question 5

The total number of balls in Box I is $6 + 9 = 15$. The total number of balls in Box II is $8 + 12 = 20$. When mixed, the total number of balls is $35$.

Calculate the basic probabilities:

$P(E_1) = \frac{15}{35} = \frac{3}{7}$

$P(E_2) = \frac{20}{35} = \frac{4}{7}$

Total red balls $F_1 = 6 + 8 = 14$, so $P(F_1) = \frac{14}{35} = \frac{2}{5}$.

Total green balls $F_2 = 9 + 12 = 21$, so $P(F_2) = \frac{21}{35} = \frac{3}{5}$.

Analyze the options:

(A) $P(E_1 \cap F_1)$ is the probability the ball is from Box I and is red, which is $\frac{6}{35}$.

Since $P(E_1) \times P(F_1) = \frac{3}{7} \times \frac{2}{5} = \frac{6}{35}$, the events are independent. Statement (A) is TRUE.

(B) $P(E_2 \cap F_2)$ is the probability the ball is from Box II and is green, which is $\frac{12}{35}$.

Since $P(E_2) \times P(F_2) = \frac{4}{7} \times \frac{3}{5} = \frac{12}{35}$, the events are independent. Statement (B) is FALSE.

(C) $P(F_1 | E_1) = \frac{6}{15} = \frac{2}{5}$ and $P(F_1 | E_2) = \frac{8}{20} = \frac{2}{5}$. They are equal. Statement (C) is TRUE.

(D) $P(F_1 | E_1) = \frac{2}{5} = 0.4$ and $P(F_2 | E_2) = \frac{12}{20} = 0.6$. $0.4$ is not greater than $0.6$. Statement (D) is FALSE.