Math Problem Statement

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.

Solution

We are tasked with finding the expectation of the sum of two ticket numbers when two tickets are drawn from a box containing 2n2n tickets. The ticket numbers are distributed in a way that nCrnCr tickets bear the number rr, where r=0,1,2,,nr = 0, 1, 2, \dots, n. Here's how we can calculate the expectation.

Step 1: Define the random variables

Let the two random variables X1X_1 and X2X_2 represent the numbers on the two tickets drawn. The total number of possible tickets is 2n2n, and we are interested in the expected value of the sum X1+X2X_1 + X_2.

Thus, the goal is to find: E(X1+X2)=E(X1)+E(X2)\mathbb{E}(X_1 + X_2) = \mathbb{E}(X_1) + \mathbb{E}(X_2) Since the tickets are drawn randomly, X1X_1 and X2X_2 are identically distributed, and: E(X1+X2)=2E(X1)\mathbb{E}(X_1 + X_2) = 2\mathbb{E}(X_1) Hence, we first calculate E(X1)\mathbb{E}(X_1), the expectation of a single ticket number.

Step 2: Calculate the expectation of X1X_1

The probability that a ticket drawn bears the number rr is proportional to how many tickets bear the number rr, which is nCrnCr. The total number of tickets is 2n2^n (since r=0nnCr=2n\sum_{r=0}^{n} nCr = 2^n).

The expectation of a single ticket number X1X_1 is given by: E(X1)=r=0nrP(X1=r)\mathbb{E}(X_1) = \sum_{r=0}^{n} r \cdot \mathbb{P}(X_1 = r) The probability of drawing a ticket with number rr is: P(X1=r)=nCr2n\mathbb{P}(X_1 = r) = \frac{nCr}{2^n} Thus, the expectation becomes: E(X1)=r=0nrnCr2n\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: E(X1)=n2\mathbb{E}(X_1) = \frac{n}{2} This follows because the sum represents the expectation of a binomial distribution B(n,12)B(n, \frac{1}{2}), whose mean is n2\frac{n}{2}.

Step 4: Calculate the expectation of the sum

Now that we have E(X1)Faulk)=n2\mathbb{E}(X_1) Faulk) = \frac{n}{2}, the expected value of the sum of the two ticket numbers is: E(X1+X2)=2n2=n\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 n\boxed{n}.

Would you like further details on any step?


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 rr followed a different distribution?
  3. How does this problem change if tickets are drawn with replacement?
  4. How would the calculation differ if nCrnCr was replaced by a different function of rr?
  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.

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Math Problem Analysis

Mathematical Concepts

Probability
Expectation
Binomial Distribution

Formulas

E(X1 + X2) = E(X1) + E(X2)
P(X1 = r) = nCr / 2^n
E(X1) = sum(r * P(X1 = r))
E(X1) for Binomial Distribution B(n, 1/2) = n/2

Theorems

Expectation Property of Identically Distributed Variables
Binomial Distribution

Suitable Grade Level

Undergraduate/Advanced High School