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Math Help - Probability with letter permutations!

  1. #1
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    Probability with letter permutations!

    13. Suppose a bag contains the letters to spell
    probability.

    a) How many four-letter arrangements are
    possible using these letters?
    I made 4 separate cases (no double letters, i doubles, b doubles, both b and i doubles) and added them to get 3074 ways. Dunno if this is correct
    b) What is the probability that Barb
    chooses four letters from the bag in the
    order that spell her name?
    2/3074

    d) What four-letter arrangement would be
    most likely to be picked? Explain your
    reasoning.
    One containing at least one i and b.

    Can anyone tell me if i did this correctly, the solutions does not have the answer for this question.
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  2. #2
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    Quote Originally Posted by ibetan View Post
    13. Suppose a bag contains the letters to spell
    probability.

    a) How many four-letter arrangements are
    possible using these letters?
    I made 4 separate cases (no double letters, i doubles, b doubles, both b and i doubles) and added them to get 3074 ways. Dunno if this is correct
    b) What is the probability that Barb
    chooses four letters from the bag in the
    order that spell her name?
    2/3074

    d) What four-letter arrangement would be
    most likely to be picked? Explain your
    reasoning.
    One containing at least one i and b.

    Can anyone tell me if i did this correctly, the solutions does not have the answer for this question.
    Ibetan,

    I get a different answer on a)-- so I suggest you post the details of your solution and then we can see where we differ.
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  3. #3
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    Lexington, MA (USA)
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    Hello, ibetan!

    I assume "arrangements" means that the order of the letters is considered.
    That is, we are spelling four-letter "words".


    13. Suppose a bag contains the letters to spell PROBABILITY.

    a) How many four-letter arrangements are possible using these letters?
    There are four cases:
    . . (1) BB and II
    . . (2) BB only
    . . (3) II only
    . . (4) No doubles



    (1) BB and II

    There are: . {4\choose2,2} \,=\,6 arrangements with \{B,B,I,I\}


    (2) BB only

    We have: . \{B,B,X,Y\}
    Choose two different letters from the other 8 letters: . {8\choose2} = 28 ways.
    Then \{B,B,x,y\} can be arranged in: . \frac{4!}{2!} \,=\,12 ways.

    Hence, there are: . 28\cdot12 \,=\,336 arrangements with BB only.


    (3) II only

    We can use the same reasoning from Case (2).

    Hence, there are: . 336 arrangements with II only.


    (4) No doubles

    Select and arrange four letters from the nine different letters.

    Hence, there are: . _9P_4 \,=\,3024 arrangements with four different letters.


    Therefore, there are: . 6 + 336 + 336 + 3924 \:=\:3702 possible arrangements.




    b) What is the probability that Barb chooses four letters from the bag
    in the order that spell her name?
    She chooses B,A,R,B in that order.

    . . \begin{array}{ccc}P(B) &=& \frac{2}{11} \\ \\[-3mm]<br />
P(A) &=& \frac{1}{10} \\ \\[-3mm]<br />
P(R) &=& \frac{1}{9} \\ \\[-3mm]<br />
P(B) &=& \frac{1}{8} \end{array}


    Therefore: . P(B,A,R,B) \;=\;\frac{2}{11}\cdot\frac{1}{10}\cdot\frac{1}{9}  \cdot\frac{1}{8} \;=\;\frac{1}{3960}




    d) What four-letter arrangement would be most likely to be picked?
    A strange question . . . It could have been worded more precisely.

    There are 3702 possible four-letter arrangements.
    . . And 3024 of them has four different letters.

    So, selecting four letters at random,
    . . you would get four different letters over 80% of the time.

    Which set of four different letters is the most likely?
    . . Um . . . all of them?

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  4. #4
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    Quote Originally Posted by awkward View Post
    Ibetan,

    I get a different answer on a)-- so I suggest you post the details of your solution and then we can see where we differ.
    Both my cases for one set of double letters were wrong.
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