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Thread: 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 $\displaystyle 11$ items that contain $\displaystyle 2$ items repeated of the first kind ('b') and $\displaystyle 2$ items repeated of the second kind ('i'). This is
    $\displaystyle \frac{11!}{2!2!}$
    (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
    $\displaystyle \frac{2!2!}{11!}$
    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 $\displaystyle 4$ items, there are $\displaystyle 9$ 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 $\displaystyle d_n$, the number of derangements of $\displaystyle n$ items, it is in the form of a recurrence relation:
    $\displaystyle d_n=nd_{n-1}+(-1)^n, n\ge 1,$ with $\displaystyle d_0$ defined as $\displaystyle d_0=1$
    You'll see that this gives:
    $\displaystyle d_1=0$
    $\displaystyle d_2=1$
    $\displaystyle d_3=2$
    $\displaystyle d_4=9$
    $\displaystyle d_5=44$
    and so on.

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

    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 $\displaystyle \frac{9}{24}$.

    Grandad
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