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Homework Help

Discussion in 'Fred's House of Pancakes' started by EricGo, May 8, 2006.

  1. EricGo

    EricGo New Member

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    Before you laugh at me, just consider most parents will reach the day that their children ask for help, and the parent flounders. My day has come ..

    You have 4 letters, and each one should go into it's addressed envelope. If the stuffing is done randomly, what is the chance of one correct, three wrong ? 2) what is the chance of *at least * one correct stuffing ?

    Please explain your thinking, not just the answer. Something more elegant than listing possible outcomes, if possible.

    Thanks much :)

    Answers have been posted below ..
     
    #1 EricGo, May 8, 2006
  2. RonH

    RonH Member

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    I'm a big believer in brute force. With only 4*3*2*1=24 combinations, you could list them and count! I believe you get 8 combinations with exactly one right which is a chance of 1/3. At least 1 right is 1-P(all wrong) = 1-9/24 = 5/8

    whoops didn't see your last condition until too late...
     
    #2 RonH, May 8, 2006
  3. ceric

    ceric New Member

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    <div class='quotetop'>QUOTE(EricGo @ May 8 2006, 08:08 PM) [snapback]251849[/snapback]</div>
    I like probability. Here it goes:

    Since the total combination is 4! = 4x3x2x1 = 24, all you need to do is count the correct combinations.
    The probability = (count of correct combinations)/24

    (1) Exactly 1 correct: pick a letter and put it into the correct envelop. The 2nd one has 2 choices (to be in wrong envelop), the 3rd one has 1 choice and so is the 4th one.
    => 2x1x1=2 combinations for each letter picked to be correct first.
    There are four such letters to begin with.
    The total combinations = 4(for 1st picking of letter) x 2 = 8 combinations.
    Therefore probability = 8/24 = 1/3

    (2) at least 1 is correct: probability = 1 - (all letters are in wrong envelops)
    - all letters in wrong envelop: 1st letter has 3 choices to be wrong, 2nd has 2 choices, 3rd and 4th have 1 choice. => 3x2x1x1=6.
    probability of all letters in wrong envelop = 6/24 = 1/4.
    Probability of at least 1 letters in correct envelop = 1-1/4 = 3/4.

    I hope this is clear and correct. No warranty comes with those answers. ;)
     
    #3 ceric, May 8, 2006
  4. TonyPSchaefer

    TonyPSchaefer Your Friendly Moderator
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    If I remember Stats class, the odds are 1:4. And then it escalates from there. The odds of randomly getting 2 correct is (1/4)x(1/4) or 1:16. Three is the same until you get to randomly getting all four correct which it 1:256 because:
    (1/4)x(1/4)x(1/4)x(1/4) =
    1:(4x4x4x4)=
    1:256

    I hope that's right.

    Having read Ceric's response, I like that one better. Maybe it's more like:
    (1/4)x(1/3)x(1/2)x(1/1) =
    1:(4x3x2x1) =
    1:24

    I did my math independantly of reviewing Ceric's final answer and now that I look and see that we got the same one, I like it.
    I also like his explination:
    You have a 1:4 chance for the first letter/envelope combination being correct.
    There's a 1:3 for the second since there are only three letters and envelopes remaining.
    There's a 1:2 for the third opportunity.
    The last opportunity is 1:1 since there's only one option.
    That's where the 1:(4x3x2x1) comes from.

    That's my final answer (until someone smarter comes along).
     
    #4 TonyPSchaefer, May 8, 2006
  5. EricGo

    EricGo New Member

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    Answers:

    1) 1/3

    2) 5/8

    That is, if you believe the back of the book.

    Ceric, I tried your approach too, but I think a fallacy is present.

    Consider problem #1 (only one correct envelope):
    Discard the first envelope you have stuffed correctly L1 in E1)
    Now E2 can hold L2, L3, or L4.
    If it holds L3, then the chance of E3 being stuffed wrong is NOT 1:2, but 1:1, because L3 has already been used. The error does not reveal itself in the first question (because it flips the probs of the last two envelopes), but shows up in #2.
     
    #5 EricGo, May 8, 2006
  6. TonyPSchaefer

    TonyPSchaefer Your Friendly Moderator
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    For the record, "smarter" includes that person who can look in the back of the book. :p
     
    #6 TonyPSchaefer, May 9, 2006
  7. priusenvy

    priusenvy Senior Member

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    I think it's justa about impossible to derive the formula for the probability of zero/at least one match without knowing the inclusion-exclusion principle.
     
    #7 priusenvy, May 9, 2006
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