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?

The correct options are:
1. The events E₁ and F₁ are independent
2. The conditional probability P(F₁|E₁) is equal to the conditional probability P(F₁|E₂)

Let us analyze the given information step-by-step:

Step 1: Summarize the contents of the boxes.
Box I contains 6 red balls and 9 green balls.
Total number of balls in Box I = 6 + 9 = 15.

Box II contains 8 red balls and 12 green balls.
Total number of balls in Box II = 8 + 12 = 20.

All balls from Box I and Box II are mixed together.
Total number of balls in total = 15 + 20 = 35.
Total red balls = 6 + 8 = 14.
Total green balls = 9 + 12 = 21.

Step 2: Define the events and calculate individual probabilities.
Let a ball be drawn at random from the 35 mixed balls.

• E₁: Event that the ball chosen belonged to Box I.
$P\left(E_1\right)=\frac{15}{35}=\frac{3}{7}$

• E₂: Event that the ball chosen belonged to Box II.
$P\left(E_2\right)=\frac{20}{35}=\frac{4}{7}$

• F₁: Event that the ball chosen is red.
$P\left(F_1\right)=\frac{14}{35}=\frac{2}{5}$

• F₂: Event that the ball chosen is green.
$P\left(F_2\right)=\frac{21}{35}=\frac{3}{5}$

Step 3: Calculate joint probabilities and conditional probabilities.

The event (E₁ ∩ F₁) means choosing a red ball from Box I.
Number of red balls from Box I = 6.
$P(E_1\cap F_1)=\frac{6}{35}$

Let's check if $P(E_1\cap F_1)=P\left(E_1\right)\times P\left(F_1\right)$:
$P\left(E_1\right)\times P\left(F_1\right)=\frac{3}{7}\times \frac{2}{5}=\frac{6}{35}$
Since $P(E_1\cap F_1)=P\left(E_1\right)P\left(F_1\right)$, the events E₁ and F₁ are independent.

Now, calculate conditional probabilities:
Conditional probability P(F₁|E₁) is the probability that the ball is red given it is from Box I:
$P\left(F_1\right|E_1)=\frac{6}{15}=\frac{2}{5}$

Conditional probability P(F₁|E₂) is the probability that the ball is red given it is from Box II:
$P\left(F_1\right|E_2)=\frac{8}{20}=\frac{2}{5}$

Thus, $P\left(F_1\right|E_1)=P\left(F_1\right|E_2)=\frac{2}{5}$.
Therefore, the conditional probability P(F₁|E₁) is equal to the conditional probability P(F₁|E₂).

Step 4: Verify remaining statements.
• For E₂ and F₂:
$P(E_2\cap F_2)=\frac{12}{35}$
$P\left(E_2\right)\times P\left(F_2\right)=\frac{4}{7}\times \frac{3}{5}=\frac{12}{35}$
Hence, E₂ and F₂ are independent (making the statement "E₂ and F₂ are dependent" FALSE).

• For conditional probabilities P(F₁|E₁) and P(F₂|E₂):
$P\left(F_1\right|E_1)=\frac{2}{5}$
$P\left(F_2\right|E_2)=\frac{12}{20}=\frac{3}{5}$
Since $\frac{2}{5}<\frac{3}{5}$, P(F₁|E₁) is less than P(F₂|E₂), making that statement FALSE.