Let n be a natural number, and there are a total of (2n+1) cards with the numbers from 0 to 2n written one by one.

A card is taken at random from among these cards, and the number written on it is X.

When Y is determined by (a) and (b) below, let {k| k∈N ,0≤k≤2n} and find the probability P(Y) that Y=k. 

(a) If X is odd, then Y= X

(b) If X is an even number containing 0, return the card and take one new card at random from all the cards and let Y be the number written on it.

To solve the problem, we need to find the probability \( P(Y = k) \) where \( Y \) is defined based on the conditions (a) and (b) given:

- If \( X \) (the number on the initially drawn card) is odd, then \( Y = X \).
- If \( X \) is even (including 0), the card is returned and a new card is drawn to determine \( Y \).

Let's break this down step by step.

1. **Determine the total number of cards and their values:**

   There are \( 2n + 1 \) cards, numbered from 0 to 2n.

2. **Probability of drawing an odd or even card:**

   - There are \( n+1 \) even cards (including 0).
   - There are \( n \) odd cards.

3. **Initial drawing probabilities:**
   
   - Probability of drawing an even number (including 0): \( \frac{n+1}{2n+1} \).
   - Probability of drawing an odd number: \( \frac{n}{2n+1} \).

4. **Determining \( Y \) when \( X \) is odd:**

   If \( X \) is odd, \( Y = X \). There are \( n \) odd numbers (1, 3, 5, ..., \( 2n-1 \)).

   - Probability \( P(Y = k) \) for an odd \( k \):
     \[
     P(Y = k) = \frac{1}{2n+1}, \quad \text{for odd } k \text{ (since any specific odd number has a probability of } \frac{1}{2n+1} \text{)}
     \]

5. **Determining \( Y \) when \( X \) is even:**

   If \( X \) is even, we return the card and draw again. In this scenario, \( Y \) is determined purely by the probability of drawing any number from the set \(\{0, 1, 2, \ldots, 2n\}\), which is equally likely for each number.

   - Probability \( P(Y = k) \) for any \( k \):
     \[
     P(Y = k) = \frac{1}{2n+1}, \quad \text{for any } k \text{ (since we are drawing a card again from the complete set)}
     \]

6. **Combining both scenarios:**

   For even \( k \):
   - \( k \) can only be the result of drawing again since \( k \) is even and not initially possible from (a).
   \[
   P(Y = k) = \frac{n+1}{2n+1} \cdot \frac{1}{2n+1} = \frac{n+1}{(2n+1)^2}, \quad \text{for even } k
   \]

   For odd \( k \):
   - \( k \) can either be from an initial odd draw or from drawing again after returning an even card.
   \[
   P(Y = k) = \frac{n}{2n+1} \cdot \frac{1}{n} + \frac{n+1}{2n+1} \cdot \frac{1}{2n+1} = \frac{1}{2n+1} + \frac{n+1}{(2n+1)^2}, \quad \text{for odd } k
   \]
   Simplifying for odd \( k \):
   \[
   P(Y = k) = \frac{2n+1}{(2n+1)^2} = \frac{1}{2n+1}, \quad \text{for odd } k
   \]

Thus, the probability \( P(Y = k) \) is:
\[
P(Y = k) = \begin{cases}
\frac{1}{2n+1} & \text{if } k \text{ is odd} \\
\frac{n+1}{(2n+1)^2} & \text{if } k \text{ is even}
\end{cases}
\]

Would you like further details or have any questions on this? 

Here are five questions you might ask next:
1. How do you compute expected values in probability?
2. Can you explain conditional probability with an example?
3. What is the probability of drawing two specific cards in succession without replacement?
4. How do you find the variance of a probability distribution?
5. What is the law of large numbers?

**Tip:** In probability problems, clearly defining each step and condition can simplify finding the solution and avoid confusion.

Explore the probability problem involving drawing cards numbered from 0 to 2n, where Y is determined by specific conditions based on whether X (the drawn card number) is odd or even. Understand how to calculate probabilities for odd and even outcomes with detailed step-by-step explanations.

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