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 chos

Step-by-step 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(E₁) = 15 35 = 3 7 P(E₂) = 20 35 = 4 7 P(F₁) = 14 35 = 2 5 P(F₂) = 21 35 = 3 5 The joint probabilities are: P(E₁ F₁) = 6 35 P(E₂ F₂) = 12 35 For statement (A): P(E₁) P(F₁) = 3 7 2 5 = 6 35 = P(E₁ F₁) Thus, E₁ and F₁ are independent. Statement (A) is TRUE. For statement (B): P(E₂) P(F₂) = 4 7 3 5 = 12 3