Question with intersects and complements.

  • Context: Undergrad 
  • Thread starter Thread starter Shawj02
  • Start date Start date
Click For Summary
SUMMARY

The discussion centers on calculating the probability of the union of two intersections, specifically P((A∩B)∪(A∩C)), given the probabilities P(A)=0.3, P(B)=0.3, and P(C)=0.7, along with P(A∩B^c)=0.2 and P(A∩C^c)=0.2. The participant concludes that since P(A∩B∩C)=0 indicates mutual exclusivity, it leads to the realization that P(A∩B)=0 and P(A∩C)=0, which initially seems incorrect. The correct formulation derived is P(A) - P(A∩B^c) + P(A) - P(A∩C^c) = P((A∩B)∪(A∩C)).

PREREQUISITES
  • Understanding of basic probability concepts, including unions and intersections.
  • Familiarity with probability notation, specifically P(A), P(A∩B), and P(A^c).
  • Knowledge of mutually exclusive events in probability theory.
  • Ability to manipulate and simplify probability expressions.
NEXT STEPS
  • Study the principles of mutually exclusive events in probability theory.
  • Learn about the inclusion-exclusion principle in probability.
  • Explore advanced probability concepts such as conditional probability and Bayes' theorem.
  • Practice solving probability problems involving unions and intersections with different scenarios.
USEFUL FOR

Students of probability theory, mathematicians, data analysts, and anyone involved in statistical analysis who seeks to deepen their understanding of probability calculations and concepts.

Shawj02
Messages
19
Reaction score
0
Ok, first post.
So I have this question, which goes something like this...
Given
P(A)=0.3, P(B)=0.3, P(C)=0.7
P(AnB^c)=0.2, P(AnC^c)=0.2, P(AnBnC)=0

Find P((AnB)U(AnC))

(Where; n =intersect, U union, ^c = complement.)

Personally my thoughts are..
P(AnBnC)=0. Therefore mutually exclusive.

And then Because probability cannot be negative. I think that leads to P(AnB)=0,P(AnC)=0 & p(BnC)=0.
Which couldn't be right, As that would give P((AnB)U(AnC)) = 0U0 = 0.

My major concern is how do I change P(AnC^c) & P(AnB^c) to something useful!

Thanks!
 
Physics news on Phys.org
Ive got
P(A)- P(AnB^c) + P(A)- P(AnC^c)= P((AnB)U(AnC))

Anyone want to double check me?
 
Shawj02 said:
Ive got
P(A)- P(AnB^c) + P(A)- P(AnC^c)= P((AnB)U(AnC))

Anyone want to double check me?


It's correct. Do you understand how? Also you obviously can get a number from it.
 
Awesome. Yeah, I understand how. Just had a block a guess.
Thanks.
 

Similar threads

Proving the Distributive Laws for Sets Using Commutative and Complement Laws
  • · Replies 6 ·
Replies
6
Views
2K
Undergrad Ways to put n balls in m boxes such that exactly k boxes are empty?
  • · Replies 29 ·
Replies
29
Views
4K
Calculate ##P(C|A')## in the given probability problem
  • · Replies 2 ·
Replies
2
Views
950
Undergrad Apparent counterexample to Cauchy-Goursat theorem (Complex Analysis)
  • · Replies 7 ·
Replies
7
Views
2K
Help needed with statistics, probability
  • · Replies 19 ·
Replies
19
Views
7K
Graduate Probability Axioms: Deriving Version 2 from Version 1
  • · Replies 3 ·
Replies
3
Views
4K
High School Unbounded Logical Trees: Constructing Natural Numbers with Logical Trees
  • · Replies 7 ·
Replies
7
Views
2K
A couple of probability homework questions
  • · Replies 1 ·
Replies
1
Views
5K
Undergrad Counterexample Required (Standard Notations)
  • · Replies 16 ·
Replies
16
Views
3K
Undergrad Can I get thresholds from logistic regression coefficients?
  • · Replies 12 ·
Replies
12
Views
2K