Alternative formula for the Rule of Addition not working for some problems

  • Context: Undergrad 
  • Thread starter Thread starter Juwane
  • Start date Start date
  • Tags Tags
    Addition Formula
Click For Summary

Discussion Overview

The discussion revolves around the application of the rule of addition in probability, specifically comparing the standard formula and an alternative formula that assumes independence. Participants explore conditions under which these formulas yield different results and the implications of independence versus dependence in probability calculations.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants note that the standard rule of addition is P(A ∪ B) = P(A) + P(B) - P(A ∩ B), which applies in all cases.
  • Others propose that the alternative formula P(A ∪ B) = P(A) + P(B) - P(A)P(B) is valid only when A and B are independent events.
  • A participant presents an example involving a shopper buying bananas and oranges, showing that the two formulas yield different results due to the dependence of the events.
  • Another participant emphasizes that when probabilities are independent, the intersection term can be omitted, leading to P(A ∪ B) = P(A) + P(B).
  • There is a clarification that the term "disjoint" refers to a different condition than "independent," highlighting a potential misunderstanding in terminology.
  • A later reply discusses the necessity of determining the intersection when probabilities are not independent, indicating that a zero value for the intersection contradicts the independence assumption.

Areas of Agreement / Disagreement

Participants generally agree that the standard formula applies in all cases, while the alternative formula is conditional on independence. However, there is disagreement regarding the implications and utility of the alternative formula, as well as the definitions of independence and disjoint events.

Contextual Notes

Participants express uncertainty about the conditions under which the alternative formula can be applied, particularly regarding the definitions of independence and disjoint events. There is also a lack of consensus on the practical utility of the alternative formula in various scenarios.

Juwane
Messages
86
Reaction score
0
In probability, the rule of addition is:

P( A \cup B) = P(A) + P(B) - P(A \cap B)

This can be written as

P( A \cup B) = P(A) + P(B) - P(A)P(B)

However, I've seen that using both formulas for the same problem gives different answers in some problems. Does anyone know why? Is there a condition for applying the second formula?
 
Physics news on Phys.org
Juwane said:
Is there a condition for applying the second formula?

Yes, the two must be independent. The first one works in all cases.
 
But consider this example:

A shopper goes to a fruit store. The probability that he will buy (1) bananas is 0.40, (2) oranges is 0.30, and (3) both bananas and oranges is 0.2. What is the probability that the shopper will buy bananas, oranges, or both?


Using the usual formula:

P( B \cup O) = P(B) + P(O) - P(B \cap O) = 0.40 + 0.30 - 0.2 = 0.50

Using the alternate formula:

P( B \cup O) = P(B) + P(O) - P(A)P(B) = 0.40 + 0.30 - (0.4)(0.3) = 0.58

Why they are different?
 
Juwane said:
Why they are different?

Because the probabilities are not independent. (Didn't you see my post?)
 
CRGreathouse said:
Because the probabilities are not independent. (Didn't you see my post?)

I did see your post. That's why I said, "But consider this example".

Anyway, when the probabilities are independent, we don't need the minus part of the formula. This is enough:

P( A \cup B) = P(A) + P(B)

Then what is the use of writing the formula in the alternate form?
 
That's disjoint, not independent.
 
Juwane said:
This is enough:

P( A \cup B) = P(A) + P(B)

Then what is the use of writing the formula in the alternate form?

Only if probabilities are independent does P(A)P(B)=P(A)\cap P(B). Otherwise you must be given or otherwise determine the intersection.

EDIT: A zero value for the intersection of P(A) and P(B) is not consistent with independence for non zero P(A),P(B).
 
Last edited:

Similar threads

  • · Replies 36 ·
2
Replies
36
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
Replies
5
Views
2K
  • · Replies 29 ·
Replies
29
Views
5K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K
Replies
1
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K