Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Statistics for CS Problem

  1. Sep 29, 2010 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant 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 [tex]\cap[/tex] 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) = [tex]\frac{P(A \cap B)}{P(B)}[/tex] provided P(B) > 0.

    3. The attempt at a solution

    I know that P([tex]\overline{A}[/tex]) = 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([tex]\overline{A}[/tex]) != P(A), then A and B are dependent. But I'm not sure if this logic is correct.
     
    Last edited: Sep 29, 2010
  2. jcsd
  3. Sep 30, 2010 #2

    statdad

    User Avatar
    Homework Helper

    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.
     
  4. Sep 30, 2010 #3
    But that's my problem, how do I use that information to get P(A | B)?
     
  5. Sep 30, 2010 #4

    statdad

    User Avatar
    Homework Helper

    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).
     
  6. Sep 30, 2010 #5
    I get 1. So P(A | B) + P(A-complement | B) = P(A|B) + .4 = 1... P(A | B) = .6?
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook