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Conditional probability problem

  1. Mar 16, 2012 #1
    1. The problem statement, all variables and given/known data

    This is in fact an example with solutions. but I don't understand the solutions. So I am here to ask for explanation.

    Details are;

    In a city there are equal number of gentleman and ladies. 10% of gentleman are regarded as "good-looking" while 10% of ladies regarded as "good-looking". People form couples randomly.

    Given that a member of a couple is good-looking, find the probability that the other member is also good-looking


    2. Relevant equations

    They first define
    G: set of good looking gentleman
    L: set of good looking ladies


    then i start to confuse here..........

    P(the other is good-looking AND a member is good-looking)
    = P((L and G)and(L Union G))
    = P(L and G)
    ..........

    P(a member is good looking)
    =P(L Union G)
    =.......

    3. The attempt at a solution




    the solution and the problem is an example of conditional probability stated in the materials
    but... i just don't understand


    In my opinion, since the first member of a couple is GIVEN to be good-looking while people form couple randomly, then it should mean that a good looking member will not have a higher or less chance of finding a good-looking member to form couple. Then why it is a conditional probability?
     
    Last edited: Mar 16, 2012
  2. jcsd
  3. Mar 16, 2012 #2

    Ray Vickson

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    Science Advisor
    Homework Helper


    Be very, very careful about wording: the question did not say "first member of a couple..." (those were *your* words), it just said "a member of a couple...". Believe it or not, that makes all the difference in the world!

    Consider a related case that might be easier to work with: say we toss two fair coins once. Outcomes are of the form HH,HT,TH,TT, where these are for coin1, coin2 in that order. If I ask: "given that the first coin shows heads, what is the probability the second coin show heads", the answer would be 1/2. If I ask "given that one coin shows heads, what is the probability that the other coin shows heads" the answer would be 1/3. A similar type of thing is occurring in this problem.

    RGV
     
  4. Mar 16, 2012 #3
    yes, but the problem is, they choose another to couple RANDOMLY. A given good looking member will not have higher or less chance coupling with another one who is either good-looking or not. Imagine a good looking member, he/she has 10% chance of coupling a good looking member too. Isn't it? if not, I really dont undesrtad what is meant by "random"
     
  5. Mar 16, 2012 #4
    I would suggest you draw a tree diagram. It is often much easier to figure these types of problems out if you have something to help you visualize.
     
  6. Mar 16, 2012 #5

    Ray Vickson

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    Science Advisor
    Homework Helper

    Go back to my two-coins example, as it is a bit easier to work with. Again, the list of possible outcomes is HH, HT, TH, TT, all equally likely. You pick one of these 4 outcomes at random, and it happens to have a H in it (that is, it is not the outcome TT). So, you must have picked either HH, HT or TH, and they are still equally likely. So, having observed an 'H' as a result of a random choice, there is now a 1/3 chance you chose HH---after all, it is the one out of three equally likely options.

    RGV
     
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