Three numbers are selected at random without replacement from the set of numbers {1, 2, 3, … n}. The conditional probability that the 3rd number lies between the 1st two. If the 1st number is known to be smaller than the 2nd, is

Options

A. `1/6`
B. `2/3`
C. `1/3`
D. `5/6`

Answer

To solve the problem, we need to find the conditional probability that the third number selected lies between the first two numbers, given that the first number is smaller than the second.

Let's denote the three selected numbers as \( X_1, X_2, \) and \( X_3 \), where:
- \( X_1 \) is the first number,
- \( X_2 \) is the second number,
- \( X_3 \) is the third number.

We are given the condition that \( X_1 < X_2 \).

### Step-by-Step Solution:

1. **Understanding the Selection**:
- We are selecting three distinct numbers from the set \( \{1, 2, 3, \ldots, n\} \).
- The total number of ways to choose 3 numbers from \( n \) numbers is given by \( \binom{n}{3} \).

2. **Arranging the Numbers**:
- Once we have selected three numbers, they can be arranged in any order. There are \( 3! = 6 \) ways to arrange three numbers.

3. **Condition Given**:
- We know that \( X_1 < X_2 \). This condition reduces the number of arrangements we need to consider.
- Given \( X_1 < X_2 \), the possible arrangements of \( X_1, X_2, \) and \( X_3 \) are:
- \( X_1, X_3, X_2 \)
- \( X_3, X_1, X_2 \)
- \( X_1, X_2, X_3 \)
- \( X_2, X_1, X_3 \)
- \( X_2, X_3, X_1 \)
- \( X_3, X_2, X_1 \)

4. **Identifying Favorable Outcomes**:
- Out of these arrangements, we need to find those where \( X_3 \) lies between \( X_1 \) and \( X_2 \).
- The arrangements that satisfy \( X_1 < X_3 < X_2 \) are:
- \( X_1, X_3, X_2 \)
- \( X_2, X_3, X_1 \)
- Thus, there are 2 favorable arrangements where \( X_3 \) is between \( X_1 \) and \( X_2 \).

5. **Calculating the Probability**:
- The total arrangements where \( X_1 < X_2 \) is 3 (since we can arrange \( X_3 \) in any of the 3 positions relative to \( X_1 \) and \( X_2 \)).
- Therefore, the conditional probability \( P \) that \( X_3 \) lies between \( X_1 \) and \( X_2 \) given \( X_1 < X_2 \) is:
\[ P(X_1 < X_3 < X_2 | X_1 < X_2) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{2}{3} \]

### Final Answer:
The conditional probability that the 3rd number lies between the 1st two, given that the 1st number is smaller than the 2nd, is \( \frac{1}{3} \).