Math Problem Statement

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.

Solution

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

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

Let's break this down step by step.

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

    There are 2n+12n + 1 cards, numbered from 0 to 2n.

  2. Probability of drawing an odd or even card:

    • There are n+1n+1 even cards (including 0).
    • There are nn odd cards.
  3. Initial drawing probabilities:

    • Probability of drawing an even number (including 0): n+12n+1\frac{n+1}{2n+1}.
    • Probability of drawing an odd number: n2n+1\frac{n}{2n+1}.
  4. Determining YY when XX is odd:

    If XX is odd, Y=XY = X. There are nn odd numbers (1, 3, 5, ..., 2n12n-1).

    • Probability P(Y=k)P(Y = k) for an odd kk: P(Y=k)=12n+1,for odd k (since any specific odd number has a probability of 12n+1)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 YY when XX is even:

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

    • Probability P(Y=k)P(Y = k) for any kk: P(Y=k)=12n+1,for any k (since we are drawing a card again from the complete set)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 kk:

    • kk can only be the result of drawing again since kk is even and not initially possible from (a). P(Y=k)=n+12n+112n+1=n+1(2n+1)2,for even kP(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 kk:

    • kk can either be from an initial odd draw or from drawing again after returning an even card. P(Y=k)=n2n+11n+n+12n+112n+1=12n+1+n+1(2n+1)2,for odd kP(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 kk: P(Y=k)=2n+1(2n+1)2=12n+1,for odd kP(Y = k) = \frac{2n+1}{(2n+1)^2} = \frac{1}{2n+1}, \quad \text{for odd } k

Thus, the probability P(Y=k)P(Y = k) is:

\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.

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Conditional Probability

Formulas

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Theorems

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Suitable Grade Level

Advanced High School or College