1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Conditional probability?

  1. Feb 24, 2016 #1
    1. The problem statement, all variables and given/known data
    suppose we have 9 balls : 2 red, 3 green, 4 yellow. and we draw 2 balls without replacement, the probability that one of them is red and the other is green is : P(R)P(G\R)+P(G)P(R\G) = (2/9)(3/8)+(3/9)(2/8)

    i faced a problem in the textbook which says: the probability that a married man watches a show is 0.4, and the prob. that a married woman watched the same show is 0.5, and the prob. the man watches this show given that his wife does is 0.7. what is the probability that a couple watches the show ?

    the answer to this problem is the intersection which is P(W)P(M\W) = 0.5*0.7 = 0.35

    my question: why didn't we treat this problem like the first one, i mean giving the answer as P(W)P(M\W)+P(M)P(W\M) ?
    2. Relevant equations
    probability of A given B = P(A\B) = P(A∩B) / P(B)

    3. The attempt at a solution
    my thinking about the solution contradiction is merged with the problem statement to be relevant
     
  2. jcsd
  3. Feb 24, 2016 #2

    andrewkirk

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    We can't use the first method in the second problem because we do not know P(W|M): the probability that the woman watches the show given that the man does. Fortunately, we can do it as above, which is even easier.

    That might lead one to ask - then why can't we do the first problem in the same way as the second one? The answer is that we can. The calculation using that approach is
    Probability of one red and one green = P(G1 | R1) P(R1) where
    G1 is the event that exactly one green is picked, and
    R1 is the event that exactly one red is picked.

    P(R1) is the prob that the first ball is red and the second isn't, plus the prob that the second ball is red and the first isn't. That's equal to:
    $$P(R1) =\left(\frac{2}{9}\cdot\frac{7}{8}+\frac{7}{9}\cdot\frac{2}{8}\right)$$
    We also have
    $$P(G1 | R1)=\frac{3}{7}$$
    Multiplying these out, you'll see it gives the same answer as the other method.

    However, doing it this way is at least as complex as the way it was done above, so no efficiency is gained by doing it this way.

    In probability there are often a number of different ways of approaching a problem. With practice, one develops intuitions about which ways are likely to be the quickest for a given sort of problem.
     
    Last edited: Feb 24, 2016
  4. Feb 24, 2016 #3
    but can't we say that this = P(M∩W) / P(M) and we already have the intersection and P(M) ? by the way the second part of the question was asking about this ( P(W|M) )
     
  5. Feb 24, 2016 #4

    andrewkirk

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No, we are not given that. The intersection is the answer to the problem, not an input to it.
    Once we have done the first part, which is to calculate P(M∩W), we can then calculate P(W|M) in the way you suggest.
     
  6. Feb 24, 2016 #5
    i may have a conflict in something, regarding this :
    is P(R∩G) = P(R)P(G\R) + P(G)P(R\G) as a whole, or is it = P(R)P(G\R) = P(G)P(R\G) as if we are adding P(R∩G) to P(R∩G) giving 2P(R∩G) ?
     
  7. Feb 24, 2016 #6

    andrewkirk

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    It's the second one
    P(R∩G)= P(R)P(G|R) = P(G)P(R|G)
    As you guess at the end of your post, the first one gives double the value of the correct answer.
     
  8. Feb 25, 2016 #7
    i think i've figured what what was wrong in my thinking about this.

    for the colored balls example we could normally treat with their tree diagram as follows:
    upload_2016-2-25_13-45-21.png

    because red, green, yellow are the all possible outcomes of this experiment, so we can normally say: probability of both green and red is : P(R)P(G|R)+P(G)P(R|G)

    while in the married man/woman problem, i was treating with the experiment like this:
    Untitled.png

    and this is wrong because man and woman are not all possible outcomes of this experiment, i mean it is not like a coin which will give you a tail or head, so P(M)P(W|M)+P(W)P(M|W) is wrong because it is based on a wrong tree, while this:
    Untitled.png
    is a right tree and can be used to get the correct values
     
    Last edited: Feb 25, 2016
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted