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Math Help - 2 Probability question

  1. #1
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    2 Probability question

    hello everyone

    i am having problem with 2 question :
    1) say i don't know the letters a-z and i need to create the word
    " Probability" from the letters "P-r-o-b-a-b-i-l-i-t-y " . ( i have exactly 11 places )

    2) i have 4 kids with different names but i don't know the names of either of them. what is the prob that i wont call either of them in the right name?

    please help me(this forum rockz )

    mormor83
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  2. #2
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    Permutations with repeated items; derangements

    Hello mormor83

    Welcome to Math Help Forum!
    Quote Originally Posted by mormor83 View Post
    hello everyone

    i am having problem with 2 question :
    1) say i don't know the letters a-z and i need to create the word
    " Probability" from the letters "P-r-o-b-a-b-i-l-i-t-y " . ( i have exactly 11 places )

    2) i have 4 kids with different names but i don't know the names of either of them. what is the prob that i wont call either of them in the right name?

    please help me(this forum rockz )

    mormor83
    1) I think the question means: The letters "p-r-o-b-a-b-i-l-i-t-y" are arranged in a random order. What is the probability that they spell the word "probability"?

    If this is what the question means, then you need to know the number of permutations (arrangements) of 11 items that contain 2 items repeated of the first kind ('b') and 2 items repeated of the second kind ('i'). This is
    \frac{11!}{2!2!}
    (For a general formula see this page.)

    Since just one of these arrangements is the correct one, the probability that this arrangement occurs at random is
    \frac{2!2!}{11!}
    2) Again, if I may re-word the question: the names of 4 children are randomly selected by each child. What is the probability that none of the children selects their own name?

    Such a selection - where no item is chosen in its 'natural' place - is called a derangement. With 4 items, there are 9 derangements. (See, for example, just here.)

    It doesn't take long to list all the possibilities with 4 items, but if you want the formula for d_n, the number of derangements of n items, it is in the form of a recurrence relation:
    d_n=nd_{n-1}+(-1)^n, n\ge 1, with d_0 defined as d_0=1
    You'll see that this gives:
    d_1=0
    d_2=1
    d_3=2
    d_4=9
    d_5=44
    and so on.

    Since the number of arrangements of 4 items is 4!\,(=24), the probability that one chosen at random is a derangement is clearly \frac{9}{24}=\frac38

    Grandad
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  3. #3
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    hello Grandad.

    thanks for the replay.I appreciate it.

    1) got it .thank u
    2) i have a little problem with the re-write:
    if i have 4 kids. each one has its own name. but i know the name of all 4 but i don't know whose name belong to who.
    the question is: if i call them what is the prob that i will miss every one of them ( be wrong every time x4 )
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  4. #4
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    Hello mormor83
    Quote Originally Posted by mormor83 View Post
    ...2) i have a little problem with the re-write:
    if i have 4 kids. each one has its own name. but i know the name of all 4 but i don't know whose name belong to who.
    the question is: if i call them what is the prob that i will miss every one of them ( be wrong every time x4 )
    Isn't that what I said? The four names are the correct names of the children, but none of them is allocated to the right child.

    For instance, if the children are A, B, C and D, one possible order could be B, C, D, A; another could be B, A, D, C. None of the letters is in its 'natural' place. That's a derangement.

    Grandad
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  5. #5
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    Hello grandad

    so that will give 9/27 possibles ways.

    thank u very much for the replay.

    mormor83
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  6. #6
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    Hello mormor83
    Quote Originally Posted by mormor83 View Post
    Hello grandad

    so that will give 9/27 possibles ways.

    thank u very much for the replay.

    mormor83
    You mean \frac{9}{24}.

    Grandad
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