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.
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You're on a roll — here are the solutions to the problems. 11. (b) A bag contains 6 blue balls, 7 green balls and 3 yellow balls of the same size. Total number of balls = 6 + 7 + 3 = 16. One ball is drawn at random from the bag and not replaced. A second ball is then drawn. 11. (b) (ii) Find the probability of obtaining two blue balls or two green balls. This means we need to find the probability of drawing two blue balls (BB) or two green balls (GG). These are mutually exclusive events. Step 1: Calculate the probability of drawing two blue balls (P(BB)). P(1st blue) = (6)/(16) After drawing one blue ball, there are 5 blue balls left and 15 total balls. P(2nd blue | 1st blue) = (5)/(15) P(BB) = (6)/(16) × (5)/(15) = (30)/(240) = (1)/(8) Step 2: Calculate the probability of drawing two green balls (P(GG)). P(1st green) = (7)/(16) After drawing one green ball, there are 6 green balls left and 15 total balls. P(2nd green | 1st green) = (6)/(15) P(GG) = (7)/(16) × (6)/(15) = (42)/(240) = (7)/(40) Step 3: Calculate the probability of obtaining two blue balls or two green balls. P(BB or GG) = P(BB) + P(GG) P(BB or GG) = (1)/(8) + (7)/(40) = (5)/(40) + (7)/(40) = (12)/(40) = (3)/(10) The probability of obtaining two blue balls or two green balls is (3)/(10). 11. (b) (iii) Find the probability of obtaining one blue ball and one green ball in any order. This means we need to find the probability of drawing a blue ball then a green ball (BG) or a green ball then a blue ball (GB). These are mutually exclusive events. Step 1: Calculate the probability of drawing a blue ball then a green ball (P(BG)). P(1st blue) = (6)/(16) After drawing one blue ball, there are 7 green balls left and 15 total balls. P(2nd green | 1st blue) = (7)/(15) P(BG) = (6)/(16) × (7)/(15) = (42)/(240) = (7)/(40) Step 2: Calculate the probability of drawing a green ball then a blue ball (P(GB)). P(1st green) = (7)/(16) After drawing one green ball, there are 6 blue balls left and 15 total balls. P(2nd blue | 1st green) = (6)/(15) P(GB) = (7)/(16) × (6)/(15) = (42)/(240) = (7)/(40) Step 3: Calculate the probability of obtaining one blue ball and one green ball in any order. P(BG or GB) = P(BG) + P(GB) P(BG or GB) = (7)/(40) + (7)/(40) = (14)/(40) = (7)/(20) The probability of obtaining one blue ball and one green ball in any order is (7)/(20). Last free one today — make it count tomorrow, or type /upgrade for unlimited.