Probability 4. A bag contains 5 red balls, 3 blue balls and 2 yellow balls. A ball is drawn and not replaced. A second ball is drawn. Find the probability of drawing: a) two red balls b) one blue ball and one yellow ball c) two yellow balls d) two balls of the same colour.

**a) Two red balls** - The total number of balls is \(10\) (5 red, 3 blue, 2 yellow). - The probability that the first ball is red is \(\frac{5}{10}\). - After drawing one red ball, there are now \(4\) red balls remaining and the total number of balls is \(9\). Thus, the probability that the second ball is red is \(\frac{4}{9}\). - Therefore, the probability of drawing two red balls is: \[ \frac{5}{10} \times \frac{4}{9} = \frac{20}{90} = \frac{2}{9}. \] --- **b) One blue ball and one yellow ball** There are two possible orders: 1. **Blue then Yellow** - The probability that the first ball is blue is \(\frac{3}{10}\). - After drawing one blue ball, there are still \(2\) yellow balls out of \(9\) balls remaining. The probability that the second ball is yellow is \(\frac{2}{9}\). - Thus, the probability for this order is: \[ \frac{3}{10} \times \frac{2}{9} = \frac{6}{90} = \frac{1}{15}. \] 2. **Yellow then Blue** - The probability that the first ball is yellow is \(\frac{2}{10}\). - After drawing one yellow ball, there are \(3\) blue balls out of \(9\) balls remaining. The probability that the second ball is blue is \(\frac{3}{9}\). - Thus, the probability for this order is: \[ \frac{2}{10} \times \frac{3}{9} = \frac{6}{90} = \frac{1}{15}. \] - Adding both probabilities together: \[ \frac{1}{15} + \frac{1}{15} = \frac{2}{15}. \] --- **c) Two yellow balls** - The probability that the first ball is yellow is \(\frac{2}{10}\). - After drawing one yellow ball, only \(1\) yellow ball remains and the total number of balls is \(9\). Thus, the probability that the second ball is yellow is \(\frac{1}{9}\). - Therefore, the probability of drawing two yellow balls is: \[ \frac{2}{10} \times \frac{1}{9} = \frac{2}{90} = \frac{1}{45}. \] --- **d) Two balls of the same colour** We consider each colour separately and then add the probabilities together. 1. **Two red balls:** \[ \frac{5}{10} \times \frac{4}{9} = \frac{20}{90}. \] 2. **Two blue balls:** \[ \frac{3}{10} \times \frac{2}{9} = \frac{6}{90}. \] 3. **Two yellow balls:** \[ \frac{2}{10} \times \frac{1}{9} = \frac{2}{90}. \] - Summing these probabilities: \[ \frac{20}{90} + \frac{6}{90} + \frac{2}{90} = \frac{28}{90} = \frac{14}{45}. \] --- **Final Answers:** a) \(\frac{2}{9}\) b) \(\frac{2}{15}\) c) \(\frac{1}{45}\) d) \(\frac{14}{45}\)