Math Problem Statement

Suppose sample space Ω = {0, 1, 2, ..., n} for some positive integer n. Let P({k}) = ck for all k ∈ Ω. Find c in terms of n (1 point) and verify that it satisfies the first and second axioms of probability (2 points).

Solution

The given problem involves a finite sample space Ω={0,1,2,,n}\Omega = \{0, 1, 2, \dots, n\} for a positive integer nn, and a probability function P({k})=ck\mathbb{P}(\{k\}) = c k for each kΩk \in \Omega, where cc is a constant to be determined. You are asked to:

  1. Find cc in terms of nn (1 point).
  2. Verify that this satisfies the first and second axioms of probability (2 points).

Step 1: Finding cc in terms of nn

The sum of the probabilities over the entire sample space must be 1 (this is the normalization condition):

k=0nP({k})=1\sum_{k=0}^{n} \mathbb{P}(\{k\}) = 1

Since P({k})=ck\mathbb{P}(\{k\}) = c k, this gives:

k=0nck=1\sum_{k=0}^{n} c k = 1

Factoring out cc:

ck=0nk=1c \sum_{k=0}^{n} k = 1

The sum of the first nn integers is a well-known formula:

k=0nk=n(n+1)2\sum_{k=0}^{n} k = \frac{n(n+1)}{2}

Substituting this into the equation:

cn(n+1)2=1c \cdot \frac{n(n+1)}{2} = 1

Solving for cc:

c=2n(n+1)c = \frac{2}{n(n+1)}

Thus, the constant cc is:

c=2n(n+1)c = \frac{2}{n(n+1)}

Step 2: Verifying the Axioms of Probability

The two fundamental axioms of probability are:

  • Axiom 1: The probability of any event is non-negative, i.e., P({k})0\mathbb{P}(\{k\}) \geq 0 for all kΩk \in \Omega.
  • Axiom 2: The sum of the probabilities of all possible outcomes is 1, i.e., k=0nP({k})=1\sum_{k=0}^{n} \mathbb{P}(\{k\}) = 1.

Verifying Axiom 1

Since P({k})=ck\mathbb{P}(\{k\}) = c k, and k0k \geq 0, the probability will always be non-negative as long as c0c \geq 0. From our earlier result, c=2n(n+1)c = \frac{2}{n(n+1)}, which is positive for any positive integer nn. Therefore, the first axiom is satisfied.

Verifying Axiom 2

We already used the second axiom in Step 1 when we summed the probabilities and set it equal to 1. Thus, the second axiom is also satisfied by the calculation:

k=0nP({k})=1\sum_{k=0}^{n} \mathbb{P}(\{k\}) = 1

Conclusion

  • The constant cc is 2n(n+1)\frac{2}{n(n+1)}.
  • The probability function satisfies both the first and second axioms of probability.

Would you like further clarification or additional details?

Here are five related questions to deepen your understanding:

  1. How would the result change if the sample space were modified?
  2. What are the implications if cc were not positive?
  3. Can this probability distribution be generalized to non-uniform cases?
  4. How would you handle the case when nn \to \infty?
  5. What if the probability P({k})\mathbb{P}(\{k\}) had a different dependence on kk?

Tip: Always check that probability distributions sum to 1, as this is a fundamental requirement in probability theory.

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Axioms of Probability
Finite Sample Space

Formulas

P({k}) = ck
Sum of first n integers: Σk = n(n+1)/2

Theorems

Axioms of Probability

Suitable Grade Level

Undergraduate Level