A box contains 2
n
tickets of which nCr

tickets bear the number r (r = 0,1,2,3, ... , n). Two tickets

are drawn from the box. Find the expectation of the sum of their numbers.

We are tasked with finding the expectation of the sum of two ticket numbers when two tickets are drawn from a box containing \(2n\) tickets. The ticket numbers are distributed in a way that \(nCr\) tickets bear the number \(r\), where \(r = 0, 1, 2, \dots, n\). Here's how we can calculate the expectation.

### Step 1: Define the random variables
Let the two random variables \( X_1 \) and \( X_2 \) represent the numbers on the two tickets drawn. The total number of possible tickets is \( 2n \), and we are interested in the expected value of the sum \( X_1 + X_2 \).

Thus, the goal is to find:
\[
\mathbb{E}(X_1 + X_2) = \mathbb{E}(X_1) + \mathbb{E}(X_2)
\]
Since the tickets are drawn randomly, \( X_1 \) and \( X_2 \) are identically distributed, and:
\[
\mathbb{E}(X_1 + X_2) = 2\mathbb{E}(X_1)
\]
Hence, we first calculate \( \mathbb{E}(X_1) \), the expectation of a single ticket number.

### Step 2: Calculate the expectation of \( X_1 \)
The probability that a ticket drawn bears the number \( r \) is proportional to how many tickets bear the number \( r \), which is \( nCr \). The total number of tickets is \( 2^n \) (since \( \sum_{r=0}^{n} nCr = 2^n \)).

The expectation of a single ticket number \( X_1 \) is given by:
\[
\mathbb{E}(X_1) = \sum_{r=0}^{n} r \cdot \mathbb{P}(X_1 = r)
\]
The probability of drawing a ticket with number \( r \) is:
\[
\mathbb{P}(X_1 = r) = \frac{nCr}{2^n}
\]
Thus, the expectation becomes:
\[
\mathbb{E}(X_1) = \sum_{r=0}^{n} r \cdot \frac{nCr}{2^n}
\]

### Step 3: Simplify the expectation expression
This summation can be simplified using the known result for the expected value of a binomial distribution:
\[
\mathbb{E}(X_1) = \frac{n}{2}
\]
This follows because the sum represents the expectation of a binomial distribution \( B(n, \frac{1}{2}) \), whose mean is \( \frac{n}{2} \).

### Step 4: Calculate the expectation of the sum
Now that we have \( \mathbb{E}(X_1) Faulk) = \frac{n}{2} \), the expected value of the sum of the two ticket numbers is:
\[
\mathbb{E}(X_1 + X_2) = 2 \cdot \frac{n}{2} = n
\]

### Final Answer:
The expected value of the sum of the numbers on the two tickets drawn is \( \boxed{n} \).

Would you like further details on any step?

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Here are some related questions to deepen your understanding:

1. How do you compute the expectation for more than two tickets drawn?
2. What if the number of tickets with number \(r\) followed a different distribution?
3. How does this problem change if tickets are drawn with replacement?
4. How would the calculation differ if \(nCr\) was replaced by a different function of \(r\)?
5. How can variance be computed for the sum of the ticket numbers?

**Tip:** The binomial distribution is often used in expectation problems because its properties allow for clean and simple calculations of mean and variance.

A box contains 2n tickets of which nCr tickets bear the number r (r = 0,1,2,3, ... , n). Two tickets are drawn from the box. Find the expectation of the sum of their numbers.

Learn how to find the expectation of the sum of two ticket numbers drawn from a box containing 2n tickets, where the numbers follow a binomial distribution. This problem involves key probability concepts like expectation and the binomial distribution. Suitable for advanced high school or undergraduate students.

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