This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
You're on a roll — Step 1: Define the probabilities of choosing each bag. A standard die has 6 faces (1, 2, 3, 4, 5, 6). Bag A is chosen if the score on the die is greater than 4 (i.e., 5 or 6). P(Bag A) = (2)/(6) = (1)/(3) Bag B is chosen if the score on the die is less than or equal to 4 (i.e., 1, 2, 3, 4). P(Bag B) = (4)/(6) = (2)/(3) Step 2: Define the contents of each bag and probabilities of drawing balls without replacement. Bag A contains 4 red (R) and 2 white (W) balls. Total = 6 balls. Bag B contains 3 red (R) and 3 white (W) balls. Total = 6 balls. Probabilities for drawing two balls from Bag A (without replacement): P(R1 from A) = (4)/(6) P(W1 from A) = (2)/(6) P(R2 from A | R1 from A) = (3)/(5) P(W2 from A | R1 from A) = (2)/(5) P(R2 from A | W1 from A) = (4)/(5) P(W2 from A | W1 from A) = (1)/(5) Probabilities for drawing two balls from Bag B (without replacement): P(R1 from B) = (3)/(6) P(W1 from B) = (3)/(6) P(R2 from B | R1 from B) = (2)/(5) P(W2 from B | R1 from B) = (3)/(5) P(R2 from B | W1 from B) = (3)/(5) P(W2 from B | W1 from B) = (2)/(5) a) Calculate the probability that one ball of each color is drawn. This means drawing one red and one white ball (RW or WR). Probability of drawing one of each color from Bag A: P(RW from A) = P(R1 from A) × P(W2 from A | R1 from A) = (4)/(6) × (2)/(5) = (8)/(30) P(WR from A) = P(W1 from A) × P(R2 from A | W1 from A) = (2)/(6) × (4)/(5) = (8)/(30) P(one of each from A) = (8)/(30) + (8)/(30) = (16)/(30) = (8)/(15) Probability of choosing Bag A and drawing one of each color: P(Bag A and one of each) = P(Bag A) × P(one of each from A) = (1)/(3) × (8)/(15) = (8)/(45) Probability of drawing one of each color from Bag B: P(RW from B) = P(R1 from B) × P(W2 from B | R1 from B) = (3)/(6) × (3)/(5) = (9)/(30) P(WR from B) = P(W1 from B) × P(R2 from B | W1 from B) = (3)/(6) × (3)/(5) = (9)/(30) P(one of each from B) = (9)/(30) + (9)/(30) = (18)/(30) = (3)/(5) Probability of choosing Bag B and drawing one of each color: P(Bag B and one of each) = P(Bag B) × P(one of each from B) = (2)/(3) × (3)/(5) = (2)/(5) Total probability of drawing one ball of each color: P(one of each) = P(Bag A and one of each) + P