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Thread: Check Answers/Need Balls Guidance

  1. #1
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    Check Answers/Need Balls Guidance

    An urn contains 6 red, 7 white and 5 blue balls. Consider the experiment of drawing 4 balls uniformly at random, with no order assigned to the balls drawn.
    Let A = "At least 3 white balls are drawn" and B = "Exactly one drawn ball is red"
    i) Calculate P(A), P(B), and P(A n B) - read as probability of the intersection of A and B.
    ii) Does knowing A has occurred increase the chances that B also has occured?
    iii) Does knowing B has occurred increase the chances that A also has occured?

    For part i) I am lost on how to get P(A). I thought about the outcomes of white and got 7/18 + 6/17 + 5/16 but this yielded a number larger than 1, and I assume this number should be relatively small.

    P(B) = .083333 I took 6/18 and divided that value by 4.

    ii) I believe A does increase the the chances of B because there are now less white balls to choose from.
    iii) I believe B does NOT increase the chances
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  2. #2
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    Re: Check Answers/Need Balls Guidance

    After a ball is picked out and checked, is it returned to the urn?
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    Re: Check Answers/Need Balls Guidance

    Assuming no replacement this is what's known as a multivariate hypergeometric distribution.

    $\displaystyle P[A] = P[w=3]+P[w=4] = \dfrac{\binom{7}{3}\left(\binom{6}{1}\binom{5}{0}+ \binom{6}{0}\binom{5}{1}\right) + \binom{7}{4}\binom{6}{0}\binom{5}{0}}{\binom{18}{4 }}=\dfrac {7}{51}$

    $P[B] = \dfrac{\binom{6}{1}\left(\binom{7}{3}\binom{5}{0}+ \binom{7}{2}\binom{5}{1}+\binom{7}{1}\binom{5}{2}+ \binom{7}{0}\binom{5}{3}\right)}{\binom{18}{4}}= \dfrac{22}{51}$

    $P[A\cap B] = P[(r,w,w,w)] = \dfrac{\binom{6}{1}\binom{7}{3} \binom{5}{0}}{ \binom{18}{4}}= \dfrac{7}{102}$

    $P[A\cap B] = P[B|A]P[A]$

    $P[B|A] = \dfrac{P[A\cap B]}{P[A]} = \dfrac{\dfrac{7}{102}}{\dfrac {7}{51}} = \dfrac 1 2$

    $P[B|A]=\dfrac 1 2 \text{ is slightly larger than } P[B] = \dfrac{22}{51}$

    $P[A|B] =\dfrac{ P[A \cap B]}{P[B]} = \dfrac{\dfrac{7}{102}}{ \dfrac{22}{51}} = \dfrac {7}{44}$

    $P[A|B] =\dfrac {7}{44} \text{ is slightly larger than } P[A] = \dfrac{7}{51}$
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    Re: Check Answers/Need Balls Guidance

    Sorry, repeat question below.
    Last edited by azollner95; May 9th 2017 at 07:43 AM.
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    Re: Check Answers/Need Balls Guidance

    Prove It, the first ball is drawn and kept out, thus not returned to the urn. Romsek, how would I interpret that data to answer parts ii) and iii)?
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    Re: Check Answers/Need Balls Guidance

    Quote Originally Posted by azollner95 View Post
    Prove It, the first ball is drawn and kept out, thus not returned to the urn. Romsek, how would I interpret that data to answer parts ii) and iii)?
    Yes and Yes.
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    Re: Check Answers/Need Balls Guidance

    SlipEternal can you briefly elaborate?
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    Re: Check Answers/Need Balls Guidance

    Quote Originally Posted by azollner95 View Post
    SlipEternal can you briefly elaborate?
    Sure:

    Quote Originally Posted by romsek View Post
    $P[B|A] = \dfrac{P[A\cap B]}{P[A]} = \dfrac{\dfrac{7}{102}}{\dfrac {7}{51}} = \dfrac 1 2$

    $P[B|A]=\dfrac 1 2 \text{ is slightly larger than } P[B] = \dfrac{22}{51}$
    So, the answer to ii) is Yes, there is an increased probability.

    Quote Originally Posted by romsek View Post
    $P[A|B] =\dfrac{ P[A \cap B]}{P[B]} = \dfrac{\dfrac{7}{102}}{ \dfrac{22}{51}} = \dfrac {7}{44}$

    $P[A|B] =\dfrac {7}{44} \text{ is slightly larger than } P[A] = \dfrac{7}{51}$
    So, the answer to iii) is yes, there is an increased probability.
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