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Basic Conditional Probability

  1. Sep 23, 2012 #1
    Bob picks 3 cats. There are 5 striped males, 6 striped females, 8 spotted males, and 4 spotted females.

    What is the prob. that Bob has atleast a striped cat given that he has picked at least a female?

    ------------------------------------------------------
    I have:

    (prob. of a striped given picked at least a female/total set) / [(Total set-prob of no females)/Total Set]=
    (???/23C3) / [(23C3-13C3)/23C3] = ???/(23C3-13C3)

    I can't figure out the conditional part though.
     
    Last edited: Sep 23, 2012
  2. jcsd
  3. Sep 23, 2012 #2

    chiro

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    Hey alexcc17.

    You might try setting up a probability table for P(Male and Striped), P(Female and Striped), P(Male and Spotted), P(Female and Spotted) as well as the rest of the table P(Male), P(Female), P(Spotted), P(Striped), and then look at what probability you need based on these quantities.
     
  4. Sep 23, 2012 #3
    We haven't learned that. Is there someway to use Baye's formula here?
     
  5. Sep 23, 2012 #4

    chiro

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    Bayes formula is based on probability statements in the form P(X), P(Y) and etc for various X,Y and so on.

    What you are doing is using some pre-defined formula, and I speculate that you are applying it without understanding it deeply enough.

    Conditional probabilities are P(A|B) = P(A and B)/P(B). If you have the table, then from that you can figure out the probabilities for these quantities and then find the final value.

    So you pick 3 cats from 4 groups with their own sizes and you can use the combinatorics to figure out what kinds of groups you get, or you can just use the probability for each selection to see how it affects the rest.

    So if you do in total you get 23C3 number of combinations, but group wise you will get a maximum of three groups picked with a minimum of 1 so the total number of group combinations you have is 4C1 + 4C2 + 4C3 = 4 + 6 + 4 = 14.

    Now from this you have to reduce these possibilities down to the set including striped cats and females (i.e. the union not intersection) and then you use the conditional formula to get the answer.

    It's basically a thing where you get sub-dividing until you get the state-space you want. We started with a lot of observations then reduced the combinations to 23C3 total possibilities with 14 group possibilities.

    Now remove the groups that can't be chosen (i.e. all males, no striped, etc) and you get a smaller number.

    After that find out how many striped possibilities are in the set, how many females and the intersection of these events, and calculate the conditional probability to get an answer.

    For future problems, it's going to be a lot easier to think about things in this way where you start with a lot of possibilities and you take your constraints to remove them one at a time so that the possibility spaces become smaller and smaller until you finally get the events that you have been looking for.

    Trying to do things with formulas that you don't understand will end up badly: if you understand them, then that's one thing but if you are just trying to match up things in an ad-hoc way it's going to do you no good.

    I know this myself because I used to do this for probability questions until I realized what the hell was going on and what I was actually doing.
     
  6. Sep 23, 2012 #5
    Wouldn't "at least 1 female" mean 10C1+10C2+10C3 because there are 10 females?

    I'm not sure I totally understand what you're saying! The part I really don't understand intuitively is the conditional part and finding what should replace the ???
     
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