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Math Help - Random Variables

  1. #1
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    Random Variables

    A special deck of 50 cards has 5 colors (red, white, blue, purple, and gold); for each color, there are cards numbered 1 through 10. 15 cards are picked at random without replacement from this deck. Let X be the number of purple cards picked. Find E(X) and Var(X).

    Heres what I did
    The probability distribution is

    P(X=0) = C(40,15) / C(50,15) picks the 15 from the 40 that are not purple.

    P(X=1) = C(40,14) *C(10,1) / C(50,15) picks 14 from 40 not purple and 1 from 10 purple cards.

    P(X=2) = C(40,13) *C(10,2) / C(50,15) picks 13 from 40 not purple and 2 from 10 purple cards.
    .
    .
    .
    .
    .
    P(X=10) = C(40,5) *C(10,10) / C(50,15) picks 5 from 40 not purple and 10 of 10 purple cards.
    I used a calculator to find the probability of each.


    Im not sure if the method i used is correct. Please comment..
    The values below are the probability.
    0 0.017871341971262
    1 0.103103895988050
    2 0.240575757305450
    3 0.297855699521034
    4 0.215688609997990
    5 0.094902988399115
    6 0.025511556021267
    7 0.004100071503418
    8 0.000372733773038
    9 0.000017053179158
    10 0.000000292340214
    Method 1
    Using the formula E(X) = sum P(U)*X(U) with the values above I got E(X) to be 3.0001
    Var(X) =E(x^2)- [E(X)]^2
    10.7143 - 3.0001^2 = 1.71
    var(x) = 1.71

    Method 2: Binomial random variable
    E(X) = np = (15) (1/5) = 3
    Var(X) = npq = (15)(1/5)(4/5) =2.4



    Im not too sure about the variance..
    Last edited by jjbrian; April 6th 2010 at 08:52 PM.
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  2. #2
    Super Member Anonymous1's Avatar
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    Quote Originally Posted by jjbrian View Post
    A special deck of 50 cards has 5 colors (red, white, blue, purple, and gold); for each color, there are cards numbered 1 through 10. 15 cards are picked at random without replacement from this deck. Let X be the number of purple cards picked. Find E(X) and Var(X).

    Heres what I did
    The probability distribution is

    P(X=0) = C(40,15) / C(50,15) picks the 15 from the 40 that are not purple.

    P(X=1) = C(40,14) *C(10,1) / C(50,15) picks 14 from 40 not purple and 1 from 10 purple cards.

    P(X=2) = C(40,13) *C(10,2) / C(50,15) picks 13 from 40 not purple and 2 from 10 purple cards.
    .
    .
    .
    .
    .
    P(X=10) = C(40,5) *C(10,10) / C(50,15) picks 5 from 40 not purple and 10 of 10 purple cards.
    I used a calculator to find the probability of each.

    Using the formula E(X) = sum P(U)*X(U)
    I got E(X) to be 3.0
    then var(X) =E(x^2)- [E(X)]^2
    to get var(x) to be 1.71

    Im not sure if the method i used is correct. Please comment..
    I didn't check your arithmetic, but your mass function, mean and variance are all done correctly.

    Good job.
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  3. #3
    MHF Contributor matheagle's Avatar
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    This is hypergeometric.
    Consider this as purple vs nonpurple.
    You can check your calculations for the mean and variance at......
    http://en.wikipedia.org/wiki/Hyperge...c_distribution
    The mean is trivial.
    The mean of a hypergeometric is the same as a binomial
    (which is the same experiment, but you replace the cards in that case).
    The variances are slightly different, but asymptotically equivalent.
    The mean is np, where n=15 and p=.2, the chance of selecting a purple.
    Hence the mean is 3.
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  4. #4
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    I changed some stuff above. I need help finding the variance. For any random variable problems, how can you tell what method you need to use? ( binomial,independent, bernoulli's, geometric, pascal,.........)
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  5. #5
    Super Member Anonymous1's Avatar
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    How did you calculate the second moment? E{X^2}
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  6. #6
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    E(X^2)= 0^2*0.017871341971262+ 1^2 * 0.103103895988050
    + 2^2 0.240575757305450 + 3^2 * 0.297855699521034
    +4^2 * 0.215688609997990
    + 5^2 * 0.094902988399115
    + 6^2 * 0.025511556021267
    + 7^2 * 0.004100071503418
    + 8^2 * 0.000372733773038
    + 9^2 * 0.000017053179158
    + 10^2 *0.000000292340214

    I calculated all this to find E(X^2) = 10.7143
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  7. #7
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    Thank you all for your help. My answers are correct.
    I did this problem the hard way

    N=50
    n=15
    m=10

    This is a hypergeometric distribution with: (Thanks to matheagle for the link )

    E(x) = nm / N = (10)(15) / 50 = 3

    V(x) = nm (N-n) (N-m) / N^2 (N-1)
    V(x) = (15)(10) (50-15)(50-10) / [ 50^2 (50-1) ]
    V(x) = (15)(10)(35)(40) / (2500)(49)
    =1.7142
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