A bowl of sweets contains 2 fruit gums, 3 mints and 5 toffees. Three sweets are chosen at random, without replacement, from the bowl. Calculate the probability that (i) the first two sweets chosen will be different. (ii) the three sweets chosen will be the same, (iii) of the three sweets chosen, the first two will be the same and the third will be a toffee, (iv) the three sweets chosen will be different.

Let’s break down the contents of the bowl: there are a total of 10 sweets (2 fruit gums + 3 mints + 5 toffees). **(i) The probability that the first two sweets chosen will be different:** To find the probability that the first two sweets are different, we can first calculate the total number of ways to choose the first two sweets from the 10. The total number of choices for the first sweet is 10, and for the second sweet to be different, there are 9 remaining choices. So we have: Total ways to pick first two sweets = \(10 \times 9 = 90\) Ways to pick two different sweets: 1. Fruit gum first (2 choices) then anything else (8 ways) = \(2 \times 8 = 16\) 2. Mint first (3 choices) then anything else (7 ways) = \(3 \times 7 = 21\) 3. Toffee first (5 choices) then anything else (6 ways) = \(5 \times 6 = 30\) Adding these together gives \(16 + 21 + 30 = 67\). Thus, the probability is: \[ P(\text{different}) = \frac{67}{90} \] --- **(ii) The probability that the three sweets chosen will be the same:** To choose three sweets of the same type, we need to consider the types with at least 3 sweets. There are no types with 3 or more (2 fruit gums, 3 mints, and 5 toffees). Thus, it's impossible to choose three sweets of the same type directly from the fruit gums, but we can take the mints or toffees: - All mints: Only 1 way to choose 3 mints. - All toffees: This is also 1 way (choosing 3 from 5 toffees). Hence, the total way to select 3 same sweets = \(1 + 1 = 2\). \[ P(\text{same}) = \frac{0 + 0 + 2}{\binom{10}{3}}=\frac{2}{120} = \frac{1}{60} \] --- **(iii) The probability that the first two will be the same and the third will be a toffee:** The first two sweets must be of the same type, and we’ll consider the possible cases: 1. **Choosing 2 mints and 1 toffee:** - Choose the mints first: \(3 \times 2 = 6\) ways (3 choices for first mint, then 2 remaining). - Then 1 toffee can be chosen from 5: \(5\) ways. - Total for this case: \(6 \times 5 = 30\). 2. **Choosing 2 toffees and 1 toffee:** - Choose the first toffee: \(5 \times 4 = 20\) ways (5 choices for first toffee, then 4 remaining). - That’s the only case since the other type (fruit gums) doesn’t qualify. So, total successful outcomes = \(30 + 20 = 50\). Thus, \[ P(\text{same first 2, third different}) = \frac{50}{120} = \frac{5}{12} \] --- **(iv) The probability that the three sweets chosen will be different:** To find this probability, we’ll calculate how many ways we can pick 3 different sweets from the various types: - Choose 3 different types from 10 sweets. We have 3 types available: fruit gum, mint, and toffee. Calculating correctly: - Choose 1 of each from the available: \(2 \times 3 \times 5\) = 30. Now we divide by the total combinations of sweets: \[ P(\text{all different}) = \frac{30}{120} = \frac{1}{4} \] In summary: (i) \( \frac{67}{90} \) (ii) \( \frac{1}{60} \) (iii) \( \frac{5}{12} \) (iv) \( \frac{1}{4} \)