Using Chebyshev's Theorem (and another minor question)

  • Thread starter Thread starter Hiche
  • Start date Start date
  • Tags Tags
    Minor Theorem
Click For Summary

Homework Help Overview

The discussion centers around the application of Chebyshev's Theorem to a discrete random variable with a specified mean and variance, specifically seeking to find an upper bound for the probability of a particular value. Additionally, there is a secondary question regarding the intersection of two events in probability theory.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to apply Chebyshev's Theorem but expresses uncertainty about selecting an appropriate value for k to establish an upper bound. They also raise a question about the intersection of two events and how to approach finding bounds for that probability.

Discussion Status

Participants are exploring different aspects of the problems presented. Some provide insights into the relationship between the union and intersection of events, while others suggest considering the implications of the sum of probabilities exceeding one. There is an ongoing examination of the bounds and conditions without reaching a definitive conclusion.

Contextual Notes

The original poster notes the existence of multiple upper bounds in the application of Chebyshev's Theorem and seeks clarification on the conditions for the intersection of events A and B, indicating a potential complexity in the problem setup.

Hiche
Messages
82
Reaction score
0

Homework Statement



If X is a discrete random variable with mean u = 12 and variance = 9, use Chebyshev's Theorem to find an upper bound for P(X = 21).

Homework Equations





The Attempt at a Solution



Now, I'm not sure about this since there are different upper bounds, right?

P(|21 - 3k < 21 < 21 + 3k|) ≤ 1 / k^2. We solve the inner inequality and we get -k < 3 < k. To find an upper bound, do we simply take a value k > 3?

And an unrelated question (just not to make another thread):

Let A and B be two events with P(A) = 3/4 and P(B) = 1/3. Explain why 1/12 ≤ P(A intersection B) ≤ 1/3. How do we approach this exactly?
 
Physics news on Phys.org
regarding your second question, just something to think about: what is P(A Union B)? what does this tell you about P(A intersect B)?
 
P(A U B) = P(A) + P(B) - P(A intersection B) = 3/4 + 1/3 - P(A intersection B) = 13/12 - P(A intersection B). So, when P(A intersection B) = 1/12, P(A U B) = 1 and when it's 1/3, P(A U B) = 3/4. Does it have to do something with that? It looks too obvious but I'm not figuring it out!
 
the fact that P(A) + P(B) > 1 tells you there has to be some overlap, right? So you're looking at kind of a best case/worst case scenario. How small could the overlap be? How large could the overlap be?
 

Similar threads

Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
862
  • · Replies 6 ·
Replies
6
Views
2K
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K