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Probability Question

  1. Mar 19, 2014 #1
    Hey, I think this is a combinatorial question and I'm too far removed from stats to remember. I thought it was originally a binomial distribution but then that didn't make sense after I calculated it.

    What is the probabilty of 2 homes with 2 cats and 1 dog and 2 homes with 3 dogs?

    How would I solve this type of question in general. Thanks.
  2. jcsd
  3. Mar 20, 2014 #2

    Simon Bridge

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    Welcome to PF;
    The question cannot be answered without more context.
    What is the known information?
  4. Mar 20, 2014 #3
    Thanks Simon,
    There are 4 cats and 8 dogs in total.
    The probability of picking a cat is simply 1/3 and the probability of picking a dog is 2/3 initially (i.e. one animal is not inherently more probable to be chosen).
    There is no replacement after picking an animal.
    Order does not matter.
    There is nothing else that I can think of. The question could be anything, 2 teams with 2 boys and 1 girl and 2 teams with 3 girls. Basically dividing the subset of the population into specific groups.
    Is there anything specific I'm not thinking about because I feel like a moron.
    I hope there is a really complex answer ... it's probably not.

    The total number of possibilities is 12C4 right?
    Last edited: Mar 20, 2014
  5. Mar 20, 2014 #4

    Simon Bridge

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    OK - so houses are twice as likely to want a dog as a cat.

    There are two homes - they take turns picking until each has three pets and each pick is independent?

    If the guess is true, then:
    You cannot run out of one animal for the selections of interest, so it does not matter how many dogs and cats are in the population: the probabilities won't change.

    Each home with combination cat-cat-dog (any order)
    What is the probability of one home picking cat-cat-dog?
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