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Math Help - probability - urgent help please

  1. #1
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    probability - urgent help please

    I'm having lots of problems with probability questions.
    Could you help me solve this? Thank you.



    4 different letters are to be sent to 4 different addresses. For each letter there is an envelope with its correct address. If the 4 letters are to be put into the 4 envelopes at random, what it the probability that ony 1 letter will be put into the envelope with its correct address?


    I thought each letter has a 1/4 chance of being put in the right envelope, and a 3/4 chance of being put in the wrong envelope, so
    P=(1/4)(3/4)(3/4)(3/4)

    what am I doing wrong?


    thank you
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  2. #2
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    Say that the letters are A,B,C, & D. Write out all 24 permutations of ABCD. You will see that there are only two cases in which A alone is in its correct place. You will also find that there only cases two having B alone in its correct place. The same is true for C&D.
    ABCD
    ABDC
    ACBD
    ACDB
    ADBC
    ADCB
    BACD
    BADC
    BCAD
    BCDA
    BDAC
    BDCA
    CABD

    CADB
    CBAD
    CBDA
    CDAB
    CDBA
    DABC
    DACB
    DBAC

    DBCA
    DCAB
    DCBA
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  3. #3
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    Thank you Plato, I got the concept. Could you please tell me how to put all this in a mathematical formula without having to list all the permutations and having to sort through them?
    Thank you so much.
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  4. #4
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    Hello, Simone!

    Four different letters are to be sent to 4 different addresses.
    For each letter there is an envelope with its correct address.
    If the 4 letters are to be put into the 4 envelopes at random,
    what is the probability that ony 1 letter will be put into the envelope with its correct address?
    There are: . 4! = {\color{blue}24} ways to place the letters.


    Suppose the letters are: . \{A,B,C,D\}
    . . and their corresponding envelopes are: . \{a,b,c,d\}

    One letter is in its correct envelope. .There are four choices for this.

    The other three must not be in their corrsponding envelopes.
    . . In how many ways can this happen?

    Suppose A is in envelope a. .Then B,C,D must be the wrong envelopes.

    There are only two ways: . \begin{array}{ccc}b&c&d\\ \hline C&D&B\\D&B&C\end{array}

    Hence, there are: . 4 \times 2 \:=\:{\color{blue}8} ways to have one letter correct.


    Therefore: . P(\text{one correct}) \;=\;\frac{8}{24} \;=\;\boxed{\frac{1}{3}}

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  5. #5
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    Quote Originally Posted by simone View Post
    Could you please tell me how to put all this in a mathematical formula without having to list all the permutations and having to sort through them.
    The answer depends on how much counting theory one knows?
    This is purely a problem in counting.

    Do you know about derangements?
    D(n) is the number of ways to rearrange a n-string in which no term is in its original position.
    D(2)=1, D(3)=2, D(4)=9 etc.
    So here fix one letter and derange the other three. 4*D(3)=8.
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  6. #6
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    thank you so much Soroban and Plato, I got it!
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