Statistics with Baye's theorem and partitions

In summary, the conversation revolved around questions and solutions for Q6, Q4, Q3, and Q1. The individuals discussed how to approach Q3 and Q1, with one person seeking clarification for Q1 and the other explaining the process for determining the distribution table. They also mentioned the possibility of including additional outcomes for Q1 to ensure the distribution adds up to one. For Q3, the concept of independence was brought up and the three possible pairs were identified as being independent. The conversation concluded with a suggestion for the values to use for P(A) and other probabilities in Q1.
  • #1
rock.freak667
Homework Helper
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Well the questions and solutions are in one for some and I will type out the rest.

Q6
http://img249.imageshack.us/img249/2757/47026197.jpg [Broken]

Q4
http://img237.imageshack.us/img237/1802/93774799.jpg [Broken]
Q3
http://img3.imageshack.us/img3/5919/34981698.jpg [Broken]

Q1
http://img19.imageshack.us/img19/7307/25729239.jpg [Broken]

I need some help doing Q3 and I need some clarification with Q1.

With Q1, if Y=2, then there are only two possibilities, (1,2) and (1,2). So X can only be equal to 3 thus I am not sure how to write out the distribution table.http://img222.imageshack.us/img222/9981/95680883.jpg [Broken]
 
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  • #2
for Q1, there's no reason not to include (2,2) as well (twice)

for Q3, three events would be independent if p(A and B and C) = p(A)p(B)p(C), and the three possible pairs are independent (e.g A and C), just a natural extension from two events
 
  • #3
Mulder said:
for Q1, there's no reason not to include (2,2) as well (twice)
Right, but wouldn't then my distribution not all add up to one either way? Σall x P(X=x)≠1

Mulder said:
for Q3, three events would be independent if p(A and B and C) = p(A)p(B)p(C), and the three possible pairs are independent (e.g A and C), just a natural extension from two events

This I know, but I don't know what numbers I'd write down for P(A) and so on.
 
  • #4
rock.freak667 said:
Right, but wouldn't then my distribution not all add up to one either way? Σall x P(X=x)≠1


You should just have Σall x P(X=x | Y=2) =1



This I know, but I don't know what numbers I'd write down for P(A) and so on.


I drew the square with corners at 0, (0,1), (1,1), (1,0) ao P(A) = P(B) =1/2 etc.
 

1. What is Baye's theorem?

Baye's theorem is a mathematical formula that describes the probability of an event occurring based on prior knowledge or information about related events. It is used to update the probability of an event as more evidence or information becomes available.

2. How is Baye's theorem used in statistics?

Baye's theorem is used in statistics to calculate conditional probabilities, which are the probabilities of an event occurring given that another event has already occurred. It is particularly useful in situations where there is uncertainty or incomplete information.

3. What is a partition in statistics?

In statistics, a partition is a division of a larger set into smaller, non-overlapping subsets. Each subset represents a distinct category or group, and the entire partition covers all possible outcomes or events. Partitions are commonly used in conjunction with Baye's theorem to calculate conditional probabilities.

4. How do you calculate probabilities using Baye's theorem and partitions?

To calculate probabilities using Baye's theorem and partitions, you first need to identify the relevant events and their probabilities. Then, you can use Baye's theorem formula to calculate the conditional probabilities by multiplying the prior probabilities with the corresponding likelihoods and dividing by the total probability across all events in the partition.

5. What are some real-life applications of Baye's theorem and partitions in statistics?

Baye's theorem and partitions have numerous applications in various fields, including medical diagnosis, spam filtering, market research, and risk assessment. For example, in medical diagnosis, Baye's theorem can be used to calculate the probability of a patient having a certain disease given their symptoms and medical history. In market research, partitions can be used to segment a customer base into different groups and calculate the likelihood of a specific group purchasing a product.

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