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Combination/Permutation question Why is my answer wrong?

  1. Dec 5, 2011 #1
    1. The problem statement, all variables and given/known data
    Four numbers are chosen independently at random, with replacement, from the set {1,2,3,4,5,6,7,8,9,10}

    Let A be the event that the four numbers are distinct.

    Let B be the event that exactly two of them are even.

    Calculate P(A) and P(B)

    2. Relevant equations



    3. The attempt at a solution

    This is easy but I dont know why im getting the worng answer..

    P(A) = (10C1).(9C1).(8C1).(7C1) / (10C1).(10C1).(10C1).(10C1) = 5040/10000 = 0.504 THIS IS CORRECT....

    Now for the dodgy one. My attempt:

    P(B) = (5C1).(5C1).(5C1).(5C1) / (10C1).(10C1).(10C1).(10C1) = 1/16 = 0.0625

    I have the solution and it tells me that P(B) is wrong.. I know what I have missed out but I feel I have done everything needed.
    The actual solution is: (4C2).(5C1).(5C1).(5C1).(5C1) / (10C1).(10C1).(10C1).(10C1) = .....

    Thanks for any help!
     
  2. jcsd
  3. Dec 5, 2011 #2

    HallsofIvy

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    I would do P(A) in a different way: The first number to be drawn can be anything at all. But in order that all 4 be distinct, the second drawn must one of the 9 remaining numbers. The probability of that is 9/10. The third number drawn must not be either of those but one of the 8 remaining numbers. The probability of that is 8/10. Finally, the fourth number drawn must not be any of those three but one of the 7 remaining numbers. The probability of that is 7/10. So the probability of those things happening is (9/10)(8/10)(7/10)= 504/1000= .504.

    As for B, the "4C2" is there because "tw0 even out of 4 numbers" can done in 4C2 different orders.

    Again, I would have done it differently: you have 10 numbers, 5 even, 5 odd so the probability of drawing an even number is 1/2. The probability of drawing "even, even, odd, odd", in that order, is [itex](1/2)^4= 1/16[/itex]. But there are 4C2= 4!/(2!)(2!)= 6 diferent "orders" or permutations of "even, even, odd, odd" so the probability of two even and 2 odd, in any order, is 6/16= 3/8.
     
  4. Dec 5, 2011 #3

    Ray Vickson

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    P{A} can be obtained more simply (as HallsofIvy has already pointed out). I would get P{B} by noting that you have two groups of numbers E={2,4,6,8,10} and O={1,3,5,7,9}, each of which has probability 1/2 of being selected at each drawing. You want the probability of getting 2E's in 4 trials, which is just like the probability of getting two heads in 4 tosses of a fair coin.

    RGV
     
  5. Dec 6, 2011 #4
    Thanks for that! I see where you guys are coming from using the nicer notation..

    Here's my next attemt.. Again I dont see where im going wrong..

    Find P(A n B) = P(A and B) = P(Distinct and exactly 2 even) ... ???

    Okay so. Here's my attempt: P(one even #) = 5/10, P(second even #) = 4/9, P(first odd #) = 5/8 since there are still 5 odd numbers remaining after taking out 2 even ones which we have chosen.. , P(second odd #) = 4/7

    Therefore P(A and B) = (1/2)(4/9)(5/8)(4/7) BUT since it can be in any order we must multiply by 4C2

    Hence: 4C2(1/2)(4/9)(5/8)(4/7) = 10/21

    which is wrong :( gahhh i get what your saying and i know this is easy.. What have i done now..

    regards jack
     
  6. Dec 6, 2011 #5

    Ray Vickson

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    P{B & A} = P{B|A}*P{A}. GIVEN A the problem looks like one of choosing without replacement, so P{B|A| = prob of 2E's in 4 samples drawn *without* replacement. You need the Hypergeometric distribution in this case.

    RGV
     
  7. Dec 6, 2011 #6

    HallsofIvy

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    In P(A and B) the event is that the four digits are distinct and exactly two of the digits are even. One way to get that is to choose the even digits first, then the two odd digits. The probability that the first digit is even is 5/10= 1/2. Given that happens, the second digit can be any of the remaining 4 even digits out of the 9 remaining digits: probability 4/9. Now we have 8 digits remaining, 3 even and 5 odd. The probability that the third digit is odd is not 5/8. Then the probability the last digit is odd is 4/7. The probability of "even, even, odd, odd" in that order is (1/2)(4/9)(5/8)(4/7).

    But now there are [itex]_4C_2= 6[/itex] different orders in which two even and two odd digits can occur so the probability is 6(1/2)(4/9)(5/8)(4/7)= 10/21
     
  8. Dec 6, 2011 #7

    vela

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    You need to choose two even digits, and you have five to choose from. How many ways can you do that? Similarly, calculate the number of ways you can select the odd digits. You should find there are 100 ways to select the four digits.

    Now take into account the number of ways you can select a particular combination in the four draws, e.g. 1234 vs. 1243 vs. 1324, etc.

    Then put it all together. How many ways can you select four digits that satisfy the criteria?
     
  9. Dec 6, 2011 #8

    Ray Vickson

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    The above is wrong. I am trying to edit or remove it, but there seems to no longer be an edit or delete button. So, here is the revision: P{B & A} = P{A}*P{B|A}. Given A the problem is similar to that giving the hypergeometric distribution, the difference being that items are replaced, but previously-drawn items are disallowed. So, for example, P{EEOO|A} = (5/10)(4/10)(5/10)(4/10), and each string of 2Es and 2Os has this same probability. So, P{B|A} = C(4,2)*(1/2)^2 *(4/10^2.

    RGV
     
  10. Dec 6, 2011 #9

    vela

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    In this approach, you're calculating the probability when the digits are selected without replacement. You were asked to find the probability when the digits are selected with replacement. So you should have P(second even) = 4/10, not 4/9, since you still have 10 digits to choose from, and so on.

    Note that this approach or the counting approach I outlined above will give you the same result. It's useful to be able to see how to solve the problem both ways. I often try to solve this type of problem multiple ways to make sure I'm not screwing it up.
     
    Last edited: Dec 6, 2011
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