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Thread: Probability Theory: Urn Problem, draw 4 balls with replacement.

  1. #1
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    Probability Theory: Urn Problem, draw 4 balls with replacement.

    An Urn Contains 1 green ball, 1 red ball, 1 yellow ball, and 1 white ball. I draw 4 balls with replacement. What is the probability that at least one color is repeated exactly twice?

    Solution:

    Let G be the event that we get exactly two balls are green, and R for red, Y for yellow, and W for white.

    We know P{G}=P{Y}=P{R}=P{W}= $\displaystyle \frac{1}{4}\cdot\frac{1}{4}\cdot\frac{3}{4}$

    So would the P{ one color being repeated exactly twice}=$\displaystyle \frac{3}{64}+\frac{3}{64}+\frac{3}{64}+\frac{3}{64 }=\frac{12}{64}$



    Note: I know this is wrong, but I just need some help to push forward on this problem.

    I know the sample space, $\displaystyle \Omega=4^4$
    I know the probability of a sample point is, $\displaystyle \omega=\frac{1}{256}$

    Does order not matter here? I am assuming it does not from how the question was asked at the end. They just want to know if it was repeated. So (G,G,Y,W) and (G,Y,W,G) are the same thing?
    Last edited by math951; Oct 2nd 2019 at 12:41 PM.
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  2. #2
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    Re: Probability Theory: Urn Problem, draw 4 balls with replacement.

    Order doesn't matter at all in this problem.

    You're neglecting all the cases where 2 colors are repeated twice.
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    Re: Probability Theory: Urn Problem, draw 4 balls with replacement.

    Quote Originally Posted by math951 View Post
    An Urn Contains 1 green ball, 1 red ball, 1 yellow ball, and 1 white ball. I draw 4 balls with replacement. What is the probability that at least one color is repeated exactly twice? Solution:
    Let G be the event that we get exactly two balls are green, and R for red, Y for yellow, and W for white.
    We know P{G}=P{Y}=P{R}=P{W}= $\displaystyle \frac{1}{4}\cdot\frac{1}{4}\cdot\frac{3}{4}$
    We have two possibilities to count.
    1) One colour is repeated twice & to other colours.
    2) Two different colours each repeated once.

    There are $\dbinom{4}{1}\cdot\dbinom{3}{2}$ to choose the content for #1. There are $\dfrac{4!}{2}$ ways to arrange that content.
    To find the probability multiply the content by arrangements and divide by $4^4$ That gives the probability of case #1.

    There are $\dbinom{4}{2}$ to choose the content for #2. There are $\dfrac{4!}{2^2}$ ways to arrange that content.

    Now post what you get.
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    Re: Probability Theory: Urn Problem, draw 4 balls with replacement.

    P{atleast one color repeated twice}= P{one color repeated exactly twice}+P{two colors repeated once}

    P{one color repeated exactly twice}= $\displaystyle \dbinom{4}{1}\cdot\dbinom{3}{2}\cdot\frac{4!}{2} \cdot\frac{1}{256}= \frac{144}{256}$
    P{two colors repeated once}=$\displaystyle \dbinom{4}{2}\cdot\frac{4!}{2^2}\cdot\frac{1}{256} =\frac{36}{256}$

    P{atleast one color repeated twice}= $\displaystyle \frac{144+36}{256}=\frac{180}{256}$
    Last edited by math951; Oct 2nd 2019 at 01:23 PM.
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    Re: Probability Theory: Urn Problem, draw 4 balls with replacement.

    Thanks Plato, but I may ask. When you think of arranging the content, how did you come to those conclusions? Is it just because 4 different balls, means 4! different ways to arrange them. where does the 2 and 2^2 come into play for the denominator in the arrangement consideration.
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    Re: Probability Theory: Urn Problem, draw 4 balls with replacement.

    Quote Originally Posted by math951 View Post
    Thanks Plato, but I may ask. When you think of arranging the content, how did you come to those conclusions? Is it just because 4 different balls, means 4! different ways to arrange them. where does the 2 and 2^2 come into play for the denominator in the arrangement consideration.
    If we arrange $GRYW$ there are $4!$ ways to do it.
    If we arrange $GGYW$ there are $\dfrac{4!}{2!}$ ways to do it.
    If we arrange $GGYY$ there are $\dfrac{4!}{(2!)^2}$ ways to do it.
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