A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 4 years ago

Probablity question: \[P(D|E) = P(DE) div P(E)\] \[P(E|D)P(D) \div P(E|D)P(D) + P(E|D^{c})P(D^{c})\] On the second line, I'm not sure 100% how we arrive at P(DE) = P(E|D)P(D). Can someone confirm it's because it's the contrapositive of P(D|E)P(E)?

  • This Question is Closed
  1. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[ P(D|E) = {P(D\cap E) \over P(E)} \]

  2. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[P(D|E) = P(DE) \div P(E) = P(E|D)P(D) \div P(E|D)P(D) + P(E|D^{c})P(D^{c})\]

  3. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    This seems latex intensive, I am scared to answer it without edit feature, sorry.

  4. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Then, what's the name of the rule/law that states that P(DE) = P(E|D)P(D)? I understand it equals P(D|E)P(E) but not he latter.

  5. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[ P(D|E) = {P(D\cap E) \over P(E)} \implies P(D\cap E) = P(D|E) \times P(E) \] again, \[ P(E|D) = {P(D\cap E) \over P(D)} \implies P(D\cap E) = P(E\cap D) = P(E|D) \times P(D) \] Hopefully, it's clear now.

  6. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    It's clear? or you need more help?

  7. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    2 seconds, doing it on paper

  8. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Sure :)

  9. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    To be somewhat pedantic, \[ P(A|B) \triangleq \frac{P(A \cap B)}{P(B)} \]

  10. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    http://en.wikipedia.org/wiki/Conditional_probability#Definition

  11. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ok, I see it now, thanks!

  12. anonymous
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Glad to help :)

  13. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.