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Math Help - Probability with marbles.

  1. #1
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    Probability with marbles.

    Hi, can someone check this and tell me if I am doing this correctly? I would really appreciate it!

    Six marbles are chosen without replacement from a box containing 10 red, 8 blue, and 3 yellow marbles. Let X be the number of blue marbles chosen.
    a) Find the probability distribution of X.
    b) Find the mean of the random variable X

    ok so for part a i did:

    P(0B) (6 choose 0) 1*(0.10)^0 (0.3)^6 = 0.000729
    P(1B) (6 choose 1) 6* (0.10)^1 (0.3)^5 = .001458
    P(2B) (6 choose 2) 15 *(0.10)^2 (0.3)^4 = .001215
    P(3B) (6 choose 3) 20 *(0.10)^3 (0.3)^3 = 0.00054
    P(4B) (6 choose 4) 15 * (0.10)^4 (0.3)^2 = 0.000135
    P(5B) (6 choose 5) 6 * (0.10)^5 (0.3)^1 = 0.00000180
    P(6B) (6 choose 6) 1 * (0.10)^6 (0.3)^0 = 0.00000010

    Is this right? Or am I doing this completely wrong? Some of the numbers I calculated aren't coming out right
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  2. #2
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    Re: Probability with marbles.

    Quote Originally Posted by ataloss55 View Post
    Six marbles are chosen without replacement from a box containing 10 red, 8 blue, and 3 yellow marbles. Let X be the number of blue marbles chosen.
    a) Find the probability distribution of X.
    b) Find the mean of the random variable X
    Let S=\dbinom{21}{6}, that is the number in the space.

    Then \mathcal{P}{(X=k)=\dfrac{\binom{8}{k}\binom{13}{6-k}}{S},~k=0,\cdots, 6
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  3. #3
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    Re: Probability with marbles.

    Ok, so i replace the k's with 0 and i got 0.031623
    I then replaced the k's with 1 and i got 0.18973

    Is this right? I am not really familiar with with equation. My teacher used a different method and i can kind of confused...
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  4. #4
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    Re: Probability with marbles.

    Where did you get the .1 and .3 from in your original post? Double check the probabilities you're using.
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