1. Problem 1: A box contains 7 fruits (4 oranges, 3 mangoes). Three fruits are drawn at random. (a) Probability all 3 are oranges with replacement: - Since after each draw the fruit is replaced, probability of orange each time is $$\frac{4}{7}$$. - Probability all three oranges: $$\left(\frac{4}{7}\right)^3 = \frac{64}{343}$$. (b) Probability all 3 are oranges without replacement: - First draw orange: $$\frac{4}{7}$$ - Second draw orange: $$\frac{3}{6}$$ (one orange removed) - Third draw orange: $$\frac{2}{5}$$ - Multiply: $$\frac{4}{7} \times \frac{3}{6} \times \frac{2}{5} = \frac{24}{210} = \frac{4}{35}$$. 2. Problem 2: Box with 6 yellow (Y),4 black (B),3 white (W) balls. Five balls picked one at a time without replacement. (i) Probability all 5 balls same color: - 5 Y: Cannot, only 6 Y available but the box has only 6 Y, so possible. - 5 B: Not possible, only 4 B available. - 5 W: Not possible, only 3 W available. - So only 5 Y possible. - Probability: Number of ways to choose 5 Y from 6 Y divided by total ways to choose any 5 from 13 balls. - Total ways: $$\binom{13}{5} = 1287$$ - Ways for 5 Y: $$\binom{6}{5} = 6$$ - Probability: $$\frac{6}{1287}$$ (ii) First 2 balls black, next 2 white, fifth yellow: - Prob 1st B: $$\frac{4}{13}$$ - Prob 2nd B: $$\frac{3}{12}$$ - Prob 3rd W: $$\frac{3}{11}$$ - Prob 4th W: $$\frac{2}{10}$$ - Prob 5th Y: $$\frac{6}{9}$$ - Multiply all: $$\frac{4}{13} \times \frac{3}{12} \times \frac{3}{11} \times \frac{2}{10} \times \frac{6}{9} = \frac{432}{154440} = \frac{3}{1075}$$ (iii) First 3 white, next 2 not white: - Prob 1st W: $$\frac{3}{13}$$ - Prob 2nd W: $$\frac{2}{12}$$ - Prob 3rd W: $$\frac{1}{11}$$ - For next 2 not white (yellow or black): total balls left 10. - Number of not white balls left: $$6 + 4 = 10$$ - Prob 4th not white: $$\frac{10}{10} = 1$$ - Prob 5th not white: $$\frac{9}{9} = 1$$ - Multiply: $$\frac{3}{13} \times \frac{2}{12} \times \frac{1}{11} \times 1 \times 1 = \frac{6}{1716} = \frac{1}{286}$$ (iv) Number of different 4-letter permutations from 'MARBLES' (7 letters, all unique): - Permutations: $$P(7,4) = \frac{7!}{(7-4)!} = 7 \times 6 \times 5 \times 4 = 840$$. 3. For parts i-v we consider the 4 letter permutations from 7 letters: (i) contains M: - Total permutations: 840 - Permutations without M: permutations of 6 letters taken 4 at a time: $$P(6,4)=6 \times5 \times4 \times3=360$$ - So containing M = $$840 - 360 = 480$$ (ii) contains E: - Same method. Total=840 - Without E: $$P(6,4) = 360$$ - Containing E = $$480$$ (iii) contains R: - Similarly: containing R = $$480$$ (iv) contains both M and R: - Choose positions for M and R (2 out of 4): $$\binom{4}{2} = 6$$ - Arrange M and R in those 2 positions: $$2! = 2$$ - Choose remaining 2 letters from remaining 5 letters: $$P(5,2) = 5 \times 4 = 20$$ - Total permutations: $$6 \times 2 \times 20 = 240$$ (v) begins with M and ends with R: - First letter fixed M, last fixed R - Middle two letters chosen from remaining 5 letters: $$P(5,2) = 5 \times 4 = 20$$ - Total permutations: $$20$$ 4. Number of ways to arrange 3 cars, 2 trailers and 1 lorry in a line: - Total objects: 6 - Since they are distinct, total permutations: $$6! = 720$$ 5. Number of ways to arrange 3 cars, 2 trailers and 1 lorry in a circle: - For circular permutations of $$n$$ distinct objects: $$(n-1)!$$ - Here $$n=6$$, so ways: $$5! = 120$$ q_count: 5