A bag contains 30 tickets numbered from 1 to 30 . Five tickets are drawn at random and arranged in ascending order . Find the probability that the third number is 20.
(ii) a bag contains 50 tickets numbered 1,2,3 ……, 50 of which five are drawn at random and arranged in ascending order of magnitude `(x_(1)ltx_(2)ltx_(3)ltx_(4)ltx_(5))` . What is the probability that `x_(3)=30`.?

Answer

To solve the given problems step by step, we will break down each part of the question.

### Part (i): Probability that the third number is 20

1. **Identify the total number of tickets**:
- There are 30 tickets numbered from 1 to 30.

2. **Calculate the total number of ways to draw 5 tickets**:
- The total number of ways to choose 5 tickets from 30 is given by the combination formula:
\[ \text{Total ways} = \binom{30}{5} \]

3. **Set conditions for the arrangement**:
- We want the third number (let's denote it as \( t_3 \)) to be 20. Therefore, we need:
- \( t_1, t_2 < 20 \) (the first two numbers must be less than 20)
- \( t_4, t_5 > 20 \) (the last two numbers must be greater than 20)

4. **Count the favorable outcomes**:
- The numbers less than 20 are from 1 to 19, giving us 19 options for \( t_1 \) and \( t_2 \).
- The numbers greater than 20 are from 21 to 30, giving us 10 options for \( t_4 \) and \( t_5 \).
- We need to choose 2 numbers from the first 19 and 2 numbers from the last 10:
\[ \text{Favorable outcomes} = \binom{19}{2} \times \binom{10}{2} \]

5. **Calculate the probability**:
- The probability \( P \) that the third number is 20 is given by:
\[ P = \frac{\text{Favorable outcomes}}{\text{Total ways}} = \frac{\binom{19}{2} \times \binom{10}{2}}{\binom{30}{5}} \]

6. **Calculate the combinations**:
- \( \binom{19}{2} = \frac{19 \times 18}{2} = 171 \)
- \( \binom{10}{2} = \frac{10 \times 9}{2} = 45 \)
- \( \binom{30}{5} = \frac{30 \times 29 \times 28 \times 27 \times 26}{5 \times 4 \times 3 \times 2 \times 1} = 142506 \)

7. **Final calculation**:
- The favorable outcomes are \( 171 \times 45 = 7695 \).
- Thus, the probability is:
\[ P = \frac{7695}{142506} \approx 0.054 \text{ (approximately)} \]

### Part (ii): Probability that \( x_3 = 30 \)

1. **Identify the total number of tickets**:
- There are 50 tickets numbered from 1 to 50.

2. **Calculate the total number of ways to draw 5 tickets**:
- The total number of ways to choose 5 tickets from 50 is:
\[ \text{Total ways} = \binom{50}{5} \]

3. **Set conditions for the arrangement**:
- We want the third number \( x_3 \) to be 30. Therefore, we need:
- \( x_1, x_2 < 30 \) (the first two numbers must be less than 30)
- \( x_4, x_5 > 30 \) (the last two numbers must be greater than 30)

4. **Count the favorable outcomes**:
- The numbers less than 30 are from 1 to 29, giving us 29 options for \( x_1 \) and \( x_2 \).
- The numbers greater than 30 are from 31 to 50, giving us 20 options for \( x_4 \) and \( x_5 \).
- We need to choose 2 numbers from the first 29 and 2 numbers from the last 20:
\[ \text{Favorable outcomes} = \binom{29}{2} \times \binom{20}{2} \]

5. **Calculate the probability**:
- The probability \( P \) that \( x_3 = 30 \) is given by:
\[ P = \frac{\text{Favorable outcomes}}{\text{Total ways}} = \frac{\binom{29}{2} \times \binom{20}{2}}{\binom{50}{5}} \]

6. **Calculate the combinations**:
- \( \binom{29}{2} = \frac{29 \times 28}{2} = 406 \)
- \( \binom{20}{2} = \frac{20 \times 19}{2} = 190 \)
- \( \binom{50}{5} = \frac{50 \times 49 \times 48 \times 47 \times 46}{5 \times 4 \times 3 \times 2 \times 1} = 2118760 \)

7. **Final calculation**:
- The favorable outcomes are \( 406 \times 190 = 77140 \).
- Thus, the probability is:
\[ P = \frac{77140}{2118760} \approx 0.0364 \text{ (approximately)} \]