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Math Help - stupid probability

  1. #1
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    stupid probability

    Each week Bob and three friends from work pool their money together and buy 12 lottery tickets that are randomly and equally divided among the group. By luck, there are two winning tickets among the 12 tickets purchased by Bob and his friends. What is the probability that different people end up with these winning tickets?
    ANS: 0.8182

    A general is planning to invade towns A, B, C and has 20 soldiers at his disposal (6 officers and 14 privates). After some thought, the general decides to select 12 soldiers to carry out the invasion and to keep the remaining 8 (and himself) behind to protect the command post. If the general takes 12 selected soldiers (3 officers and 9 privates), and randomly selects 4 soldiers w/o replacement, to invade towns A, B, C, what is the probability that exactly one officer ends up being sent to each town?
    ANS: 0.2909
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  2. #2
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    Quote Originally Posted by skhan
    Each week Bob and three friends from work pool their money together and buy 12 lottery tickets that are randomly and equally divided among the group. By luck, there are two winning tickets among the 12 tickets purchased by Bob and his friends. What is the probability that different people end up with these winning tickets?
    ANS: 0.8182
    I did this problem a got a result different from yours maybe I am missing something.

    A person get 3 tickets. Let us find the probability that he gets 2 winning tickets out of the 3. Meaning 2 good 1 bad. This is,
    \frac{_3C_2\cdot _{10}C_1}{_{12}C_3}=\frac{3}{22}
    Thus, the probability of this NOT happening, which is what you are trying to answer is,
    \frac{19}{22}\approx .8636.
    It does not match with your answer.
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  3. #3
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    Quote Originally Posted by skhan
    ... and randomly selects 4 soldiers w/o replacement, to invade towns A, B, C, what is the probability that exactly one officer ends up being sent to each town?
    0%??
    If it is done without replacement then how can the general sent the same soldier again?
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  4. #4
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    A general is planning to invade towns A, B, C and has 20 soldiers at his disposal (6 officers and 14 privates). After some thought, the general decides to select 12 soldiers to carry out the invasion and to keep the remaining 8 (and himself) behind to protect the command post. If the general takes 12 selected soldiers (3 officers and 9 privates), and randomly selects 4 soldiers w/o replacement, to invade towns A, B, C, what is the probability that exactly one officer ends up being sent to each town?
    ANS: 0.2909

    i dont think hes sending the same soldier again because in the original 12 he has 3 officers and 9 privares and then he choose 4 soldiers from the original 12 which could be offciers and/or privates...right?
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  5. #5
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    Az for the first problem I made a mistake in computate.

    Here is the basic idea. Find the probability that a person gets two winning tickets. Then from 1 subtract this result. This will tell you the propability of a person NOT getting two winning tickets which you are trying to find.

    There are:
    4 People
    12 Tickets
    Each person gets 3 tickets
    There are 2 winning tickets 10 losing tickets.

    The probability that any given person gets two tickets is:
    1)There are _2C_2=1 ways of chosing 2 winning tickets and _{10}C_1=10 of choosing a losing ticket. In total the number of ways choosing two winning tickets and a losing is:
    _2C_2\cdot _{10}C_1=10
    2)There are _{12}C_3=220 ways of chosing 3 tickets out of the 12.
    3)Thus, the probability is for a selected person is \frac{1}{22}

    Now, to find the probability that any person gets a double win is to multiply by 4 because in the case above we only considered a case with a specific person. Thus, probability is \frac{4}{22}=\frac{2}{11}

    Thus, the probability of this NOT happening is,
    \frac{9}{11}\approx .8181
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  6. #6
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    Quote Originally Posted by skhan
    A general is planning to invade towns A, B, C and has 20 soldiers at his disposal (6 officers and 14 privates). After some thought, the general decides to select 12 soldiers to carry out the invasion and to keep the remaining 8 (and himself) behind to protect the command post. If the general takes 12 selected soldiers (3 officers and 9 privates), and randomly selects 4 soldiers w/o replacement, to invade towns A, B, C, what is the probability that exactly one officer ends up being sent to each town?
    ANS: 0.2909

    i dont think hes sending the same soldier again because in the original 12 he has 3 officers and 9 privares and then he choose 4 soldiers from the original 12 which could be offciers and/or privates...right?
    The first part of the problem is irrelevent.
    I understand the question like this, i think

    There are 12 soldiers - 3 officers and 9 privates
    The general randomly selectes 4 people and sents them.
    Then from the remaining 8 he selects and he sents.
    Then from the remaining 4 he sends them all.

    The question is what is probability that an exactly one officer was sent to A or B or C.
    --------
    Try it like this,
    write a box which shows all possible distributions of officers which is:
    Code:
    0  0  4
    0  1  3
    0  2  2
    0  3  1
    0  4  0
    1  3  0
    2  2  0
    3  1  0
    4  0  0
    1  1  2
    1  2  1
    2  1  1
    Which one(s) fit your question?
    Last edited by ThePerfectHacker; May 9th 2006 at 02:48 PM.
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  7. #7
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    There are 12 soldiers - 3 officers and 9 privates
    The general randomly selectes 4 people and sents them.
    Then from the remaining 8 he selects and he sents.
    Then from the remaining 4 he sends them all.
    im getting a little confused by what you wrote here
    - there are 12 soldiers which consist of 3 officers and 9 privates and of those 12 soldiers 4 are selected at random

    so shouldnt we find the probability that all 3 officers (for the 4 soldiers that will invade) will be selected from the oiriginal 12 since each needs to invade towns A, B, and C
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