A bag contains 6 white balls and 4 black balls. Two balls are drawn one after another with replacement. a) If A and B are mutually exclusive events, then write the principle of P(A U B). b) Present th

You're on a roll — here are the solutions to problem 6: Given: Number of white balls (W) = 6 Number of black balls (B) = 4 Total number of balls = 6 + 4 = 10 Two balls are drawn one after another with replacement. a) If A and B are mutually exclusive events, then write the principle of P(A U B). If A and B are mutually exclusive events, they cannot occur at the same time, meaning their intersection is empty, P(A B) = 0. The principle for the probability of their union is: P(A B) = P(A) + P(B) The final answer is P(A B) = P(A) + P(B). b) Present the probability of all the possible outcomes of given event in a tree diagram. Step 1: Calculate probabilities for the first draw. P(White on 1st draw) = (6)/(10) P(Black on 1st draw) = (4)/(10) Step 2: Calculate probabilities for the second draw. Since the drawing is with replacement, the probabilities for the second draw are the same as the first. If the first ball is white: P(White on 2nd draw | 1st is White) = (6)/(10) P(Black on 2nd draw | 1st is White) = (4)/(10) If the first ball is black: P(White on 2nd draw | 1st is Black) = (6)/(10) P(Black on 2nd draw | 1st is Black) = (4)/(10) Step 3: List all possible outcomes and their probabilities. • Outcome (White, White): P(W,W) = (6)/(10) × (6)/(10) = (36)/(100) • Outcome (White, Black): P(W,B) = (6)/(10) × (4)/(10) = (24)/(100) • Outcome (Black, White): P(B,W) = (4)/(10) × (6)/(10) = (24)/(100) • Outcome (Black, Black): P(B,B) = (4)/(10) × (4)/(10) = (16)/(100) The tree diagram would visually represent these steps: Start with a single point. Branch 1: First Draw • Branch for White (probability 6/10) • Branch for Black (probability 4/10) From each of these branches, create a second set of branches for the Second Draw: • From First White branch: • Branch for Second White (probability 6/10) -> Outcome (W,W) with probability 36/100 • Branch for Second Black (probability 4/10) -> Outcome (W,B) with probability 24/100 • From First Black branch: • Branch for Second White (probability 6/10) -> Outcome (B,W) with probability 24/100 • Branch for Second Black (probability 4/10) -> Outcome (B,B) with probability 16/100 c) Find the probability of getting same coloured ball? Step 1: Identify outcomes for same colored balls. The outcomes for getting same colored balls are (White, White) and (Black, Black). Step 2: Sum their probabilities. P(Same color) = P(W,W) + P(B,B) P(Same color) = (36)/(100) + (16)/(100) P(Same color) = (52)/(100) Step 3: Simplify the fraction. P(Same color) = (13)/(25) The final answer is (13)/(25). d) Find the ratio of the probabilities of getting same coloured balls and different colored balls. Step 1: Calculate the probability of getting different colored balls. The outcomes for getting different colored balls are (White, Black) and (Black, White). P(Different color) = P(W,B) + P(B,W) P(Different color) = (24)/(100) + (24)/(100) P(Different color) = (48)/(100) Step 2: Form the ratio of P(Same color) to P(Different color). Ratio = P(Same color) : P(Different color) Ratio = (52)/(100) : (48)/(100) Step 3: Simplify the ratio. Ratio = 52 : 48 Divide both sides by their greatest common divisor, which is 4. Ratio = (52)/(4) : (48)/(4) Ratio = 13 : 12 The final answer is 13:12. What's next?