Are Events A and B Independent Based on Given Probabilities?

  • Thread starter Thread starter frankfjf
  • Start date Start date
  • Tags Tags
    Cs Statistics
Click For Summary

Homework Help Overview

The discussion revolves around determining the independence of two events, A and B, based on given probabilities. The original poster presents the probabilities P(A) = 0.6 and P(¬A | B) = 0.4, and seeks to verify if A and B are independent using these values.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand how to derive P(A | B) from the given conditional probability P(¬A | B). Some participants suggest that it should be straightforward to obtain P(A | B) from the provided information.

Discussion Status

Participants are engaged in exploring the relationship between the probabilities and how they relate to the concept of independence. There is an ongoing inquiry into the correct application of the definitions and relationships between the probabilities, with some guidance being offered on how to manipulate the probabilities to find P(A | B).

Contextual Notes

The original poster expresses uncertainty about the logic behind determining independence and the professor's indication that the solution should be brief adds a constraint to their approach. There is also a mention of difficulties with the notation in LaTeX, which may affect clarity in communication.

frankfjf
Messages
166
Reaction score
0

Homework Statement



Suppose P(A) = .6, P(\overline{A} | B) = .4. Check whether events A and B are independent.

Homework Equations



Two events A and B are said to be independent if P(A|B) = P(A). This is equivalent to stating that P(A \cap B) = P(A)P(B)

If A and B are any two events, then the conditional probability of A given B, denoted by P(A | B), is P(A | B) = \frac{P(A \cap B)}{P(B)} provided P(B) > 0.

The Attempt at a Solution



I know that P(\overline{A}) = 1 - P(A) = .4

However, I'm not sure how to use the information given to check for independence. The professor says the solution should be very brief, but it's not coming to me.

Thinking on it a little more and looking it up online, it would seem that if P(A|B) = P(A) means A and B are independent, but P(\overline{A}) != P(A), then A and B are dependent. But I'm not sure if this logic is correct.
 
Last edited:
Physics news on Phys.org
You have P(A-complement | B) - it should be very simple to obtain P(A | B) from it.

Note: couldn't get the \bar{A} construct to work in Latex.
 
But that's my problem, how do I use that information to get P(A | B)?
 
frankfjf said:
But that's my problem, how do I use that information to get P(A | B)?

What do you get when you add P(A) to P(A-complement)? the same result works with P(A |B) and P(A-complement | B).
 
I get 1. So P(A | B) + P(A-complement | B) = P(A|B) + .4 = 1... P(A | B) = .6?
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
3K
Replies
23
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 8 ·
Replies
8
Views
4K