A class consists of 10 boys and 8 girls. Three stuents are selected at random. Find the probability that the selected group has:
(i) all boys (ii) all girls (iii) 2 boys and 1 girl.

Answer

To solve the problem, we need to find the probability of selecting groups of students from a class consisting of 10 boys and 8 girls. We will calculate the probabilities for three scenarios: (i) all boys, (ii) all girls, and (iii) 2 boys and 1 girl.

### Step-by-Step Solution:

**Step 1: Calculate the total number of students.**
- Total students = Number of boys + Number of girls = 10 + 8 = 18.

**Step 2: Calculate the total ways to choose 3 students from 18.**
- The total number of ways to choose 3 students from 18 is given by the combination formula \( C(n, r) = \frac{n!}{r!(n-r)!} \).
- Thus, \( \text{Total ways} = C(18, 3) = \frac{18!}{3!(18-3)!} = \frac{18 \times 17 \times 16}{3 \times 2 \times 1} = 816 \).

**Step 3: Calculate the probability of selecting all boys.**
- The number of ways to choose 3 boys from 10 is \( C(10, 3) \).
- Thus, \( C(10, 3) = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \).
- The probability of selecting all boys is:
\[ P(\text{All Boys}) = \frac{C(10, 3)}{C(18, 3)} = \frac{120}{816} = \frac{5}{34}. \]

**Step 4: Calculate the probability of selecting all girls.**
- The number of ways to choose 3 girls from 8 is \( C(8, 3) \).
- Thus, \( C(8, 3) = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \).
- The probability of selecting all girls is:
\[ P(\text{All Girls}) = \frac{C(8, 3)}{C(18, 3)} = \frac{56}{816} = \frac{7}{102}. \]

**Step 5: Calculate the probability of selecting 2 boys and 1 girl.**
- The number of ways to choose 2 boys from 10 is \( C(10, 2) \) and the number of ways to choose 1 girl from 8 is \( C(8, 1) \).
- Thus, \( C(10, 2) = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45 \) and \( C(8, 1) = 8 \).
- Therefore, the total ways to choose 2 boys and 1 girl is:
\[ C(10, 2) \times C(8, 1) = 45 \times 8 = 360. \]
- The probability of selecting 2 boys and 1 girl is:
\[ P(2 \text{ Boys and } 1 \text{ Girl}) = \frac{C(10, 2) \times C(8, 1)}{C(18, 3)} = \frac{360}{816} = \frac{15}{34}. \]

### Final Answers:
- (i) Probability of all boys: \( \frac{5}{34} \)
- (ii) Probability of all girls: \( \frac{7}{102} \)
- (iii) Probability of 2 boys and 1 girl: \( \frac{15}{34} \)