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    From June 2012 S1 AQA A level....
    Q) Three properties are selected at random from those on the estate which have exactly
    3 bedrooms.
    Calculate the probability that one property has 2 toilets, one has 3 toilets and the
    other has at least 4 toilets. Give your answer to three decimal places.
    I got P(2T  3T  ≥4T | B = 3) = 72/194 x 99/193 x 16/192.
    I dont understand why you now have to times this by 6 or 3! as mark scheme says. Could someone please explain why you times by this and tell me how to know what to times by on these type of questions as i often slip up on knowing what to times by.
    Thanks.
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    (Original post by redmcq)
    From June 2012 S1 AQA A level....
    Q) Three properties are selected at random from those on the estate which have exactly
    3 bedrooms.
    Calculate the probability that one property has 2 toilets, one has 3 toilets and the
    other has at least 4 toilets. Give your answer to three decimal places.
    I got P(2T  3T  ≥4T | B = 3) = 72/194 x 99/193 x 16/192.
    I dont understand why you now have to times this by 6 or 3! as mark scheme says. Could someone please explain why you times by this and tell me how to know what to times by on these type of questions as i often slip up on knowing what to times by.
    Thanks.
    Tt's all to do with permutations. If you changed your problem (so that it'd be the same, but with easier things to picture) to have red, blue and green shapes (and multiple of each lots, but it doesn't matter how many) in a bag and I asked you the probability of finding 1 red, 1 blue and 1 green, then there are 6 possible combinations. RBG, RGB, GBR, GRB, BGR and BRG. And the probability of each one is just like finding what you did, 72/194 * 99/193 * 16/192 so you multiply it by 3!.

    So you may think.. is there a formula for finding what you multiply by instead of finding all the combinations? And there is.

    If you want to sort n objects, with of type 1, of type 2.... and then the formula is and in your problem, k1 = k2 = k3 = 1 so the number is just 3! / 1!1!1! = 6.
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    (Original post by SeanFM)
    Tt's all to do with permutations. If you changed your problem (so that it'd be the same, but with easier things to picture) to have red, blue and green shapes (and multiple of each lots, but it doesn't matter how many) in a bag and I asked you the probability of finding 1 red, 1 blue and 1 green, then there are 6 possible combinations. RBG, RGB, GBR, GRB, BGR and BRG. And the probability of each one is just like finding what you did, 72/194 * 99/193 * 16/192 so you multiply it by 3!.

    So you may think.. is there a formula for finding what you multiply by instead of finding all the combinations? And there is.

    If you want to sort n objects, with of type 1, of type 2.... and then the formula is and in your problem, k1 = k2 = k3 = 1 so the number is just 3! / 1!1!1! = 6.
    Thanks for the help! im a bit confused to as what the K1 , K2 etc is and what it represents. what does the k represent in my example also? Thanks
 
 
 
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