The problem describes drawing two balls without replacement from a bag containing 7 red (R), 4 white (W), and 6 blue (B) balls, for a total of 7+4+6=17 balls.

Welcome back Lonare — missed you this week. Here are the solutions for the probability problem: a) Represent the information above on a tree diagram The problem describes drawing two balls without replacement from a bag containing 7 red (R), 4 white (W), and 6 blue (B) balls, for a total of 7+4+6=17 balls. The tree diagram shows the probabilities for the first draw and the conditional probabilities for the second draw. First Draw: • P(R_1) = (7)/(17) • P(W_1) = (4)/(17) • P(B_1) = (6)/(17) Second Draw (without replacement): • If R_1: P(R_2|R_1) = (6)/(16), P(W_2|R_1) = (4)/(16), P(B_2|R_1) = (6)/(16) • If W_1: P(R_2|W_1) = (7)/(16), P(W_2|W_1) = (3)/(16), P(B_2|W_1) = (6)/(16) • If B_1: P(R_2|B_1) = (7)/(16), P(W_2|B_1) = (4)/(16), P(B_2|B_1) = (5)/(16) The tree diagram in the image correctly illustrates these probabilities. b) Use the tree diagram to find the probability that: i) The second ball drawn is red This occurs if the sequence of draws is (Red, Red), (White, Red), or (Blue, Red). Step 1: Identify the probabilities for each path. P(R_1 and R_2) = (7)/(17) × (6)/(16) = (42)/(272) P(W_1 and R_2) = (4)/(17) × (7)/(16) = (28)/(272) P(B_1 and R_2) = (6)/(17) × (7)/(16) = (42)/(272) Step 2: Sum these probabilities. P(2nd ball is Red) = (42)/(272) + (28)/(272) + (42)/(272) = (42+28+42)/(272) = (112)/(272) Step 3: Simplify the fraction. P(2nd ball is Red) = (112 ÷ 16)/(272 ÷ 16) = (7)/(17) The probability that the second ball drawn is red is (7)/(17). ii) The two balls are of the same colour This occurs if the sequence of draws is (Red, Red), (White, White), or (Blue, Blue). Step 1: Identify the probabilities for each path. P(R_1 and R_2) = (7)/(17) × (6)/(16) = (42)/(272) P(W_1 and W_2) = (4)/(17) × (3)/(16) = (12)/(272) P(B_1 and B_2) = (6)/(17) × (5)/(16) = (30)/(272) Step 2: Sum these probabilities. P(same colour) = (42)/(272) + (12)/(272) + (30)/(272) = (42+12+30)/(272) = (84)/(272) Step 3: Simplify the fraction. P(same colour) = (84 ÷ 4)/(272 ÷ 4) = (21)/(68) The probability that the two balls are of the same colour is (21)/(68). iii) No white ball is drawn This means both balls drawn are either red or blue. The possible sequences are (Red, Red), (Red, Blue), (Blue, Red), or (Blue, Blue). Step 1: Identify the probabilities for each path. P(R_1 and R_2) = (7)/(17) × (6)/(16) = (42)/(272) P(R_1 and B_2) = (7)/(17) × (6)/(16) = (42)/(272) P(B_1 and R_2) = (6)/(17) × (7)/(16) = (42)/(272) P(B_1 and B_2) = (6)/(17) × (5)/(16) = (30)/(272) Step 2: Sum these probabilities. P(no white ball) = (42)/(272) + (42)/(272) + (42)/(272) + (30)/(272) = (42+42+42+30)/(272) = (156)/(272) Step 3: Simplify the fraction. P(no white ball) = (156 ÷ 4)/(272 ÷ 4) = (39)/(68) The probability that no white ball is drawn is (39)/(68). iv) At least one ball is blue It is easier to calculate the complement: 1 - P(no blue ball). "No blue ball" means both balls drawn are either red or white. The possible sequences are (Red, Red), (Red, White), (White, Red), or (White, White). Step 1: Identify the probabilities for each path where no blue ball is drawn. P(R_1 and R_2) = (7)/(17) × (6)/(16) = (42)/(272) P(R_1 and W_2) = (7)/(17) × (4)/(16) = (28)/(272) P(W_1 and R_2) = (4)/(17) × (7)/(16) = (28)/(272) P(W_1 and W_2) = (4)/(17) × (3)/(16) = (12)/(272) Step 2: Sum these probabilities to find P(no blue ball). P(no blue ball) = (42)/(272) + (28)/(272) + (28)/(272) + (12)/(272) = (42+28+28+12)/(272) = (110)/(272) Step 3: Simplify the fraction. P(no blue ball) = (110 ÷ 2)/(272 ÷ 2) = (55)/(136) Step 4: Calculate P(at least one blue ball). P(at least one blue ball) = 1 - P(no blue ball) = 1 - (55)/(136) = (136)/(136) - (55)/(136) = (136-55)/(136) = (81)/(136) The probability that at least one ball is blue is (81)/(136). Drop the next question! 📸