19. A box contains three blue bulbs, four goen bulbs and five red bulbs. Four bulbs ars taken out of the box at random and without replacements. What is the probability that (i) all the four bulbs are of the same colour; (ii) the first two are of the same colour and the last two are of different colours.

To solve this problem, we need to calculate the probabilities based on the given conditions. ### Known Conditions: - Total bulbs: 3 blue + 4 green + 5 red = 12 bulbs - We are drawing 4 bulbs without replacement. ### Total Ways to Choose 4 Bulbs: The total number of ways to choose 4 bulbs from 12 is given by the combination formula: \[ \text{Total ways} = \binom{n}{r} = \frac{n!}{r!(n-r)!} \] Where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose. So, the total ways to choose 4 bulbs from 12 is: \[ \text{Total ways} = \binom{12}{4} \] ### (i) Probability that all four bulbs are of the same colour: We need to consider the cases for each colour: 1. **All Blue**: We can choose 4 blue bulbs, but there are only 3 blue bulbs available. So, this case is not possible. 2. **All Green**: We can choose 4 green bulbs from 4 available. \[ \text{Ways to choose 4 green} = \binom{4}{4} = 1 \] 3. **All Red**: We can choose 4 red bulbs from 5 available. \[ \text{Ways to choose 4 red} = \binom{5}{4} = 5 \] Now, we can sum the successful outcomes: \[ \text{Total successful outcomes} = 0 + 1 + 5 = 6 \] Now, we can calculate the probability: \[ P(\text{all same colour}) = \frac{\text{Total successful outcomes}}{\text{Total ways}} = \frac{6}{\binom{12}{4}} \] ### (ii) Probability that the first two are of the same colour and the last two are of different colours: We can break this down into cases based on the colour of the first two bulbs. 1. **First two blue**: The last two must be one green and one red. - Ways to choose 2 blue: \(\binom{3}{2} = 3\) - Ways to choose 1 green: \(\binom{4}{1} = 4\) - Ways to choose 1 red: \(\binom{5}{1} = 5\) - Total for this case: \(3 \times 4 \times 5 = 60\) 2. **First two green**: The last two must be one blue and one red. - Ways to choose 2 green: \(\binom{4}{2} = 6\) - Ways to choose 1 blue: \(\binom{3}{1} = 3\) - Ways to choose 1 red: \(\binom{5}{1} = 5\) - Total for this case: \(6 \times 3 \times 5 = 90\) 3. **First two red**: The last two must be one blue and one green. - Ways to choose 2 red: \(\binom{5}{2} = 10\) - Ways to choose 1 blue: \(\binom{3}{1} = 3\) - Ways to choose 1 green: \(\binom{4}{1} = 4\) - Total for this case: \(10 \times 3 \times 4 = 120\) Now, we can sum the successful outcomes: \[ \text{Total successful outcomes} = 60 + 90 + 120 = 270 \] Now, we can calculate the probability: \[ P(\text{first two same, last two different}) = \frac{\text{Total successful outcomes}}{\text{Total ways}} = \frac{270}{\binom{12}{4}} \] ### Final Calculation: Now we need to calculate \(\binom{12}{4}\): \[ \binom{12}{4} = \frac{12!}{4!(12-4)!} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495 \] Now we can substitute this value into our probabilities: 1. For (i): \[ P(\text{all same colour}) = \frac{6}{495} = \frac{2}{165} \] 2. For (ii): \[ P(\text{first two same, last two different}) = \frac{270}{495} = \frac{54}{99} = \frac{18}{33} = \frac{6}{11} \] ### Final Answers: (i) The probability that all four bulbs are of the same colour is \(\frac{2}{165}\). (ii) The probability that the first two are of the same colour and the last two are of different colours is \(\frac{6}{11}\).