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Thread: More Urns and Balls

  1. #1
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    Exclamation More Urns and Balls

    An Urn contains 5 red, 4 white, and 3 green balls. The experiment E is as follows: draw 3 balls (at random) one at a time, WITH replacement of the ball in the urn after each draw. Each outcome is an ordered triplet of the form (First ball, Second ball, Third Ball).
    Let A: "The second and third balls are the same color"
    Let B: "The first and third balls are the same color?
    Let S be the sample space.

    i) Find P(S), P(A), P(B), and P(A n B) - read as probability of intersection of A and B.
    ii) Does knowing B has occurred increase the chances that A has also occurred?

    i) For P(S) I got 1. I am stuck on the rest of the probabilities
    ii) I believe the answer is No, but I am not sure how true my logic is. I feel like since there is replacement, other events already occurring does not increase the likelihood of other events.
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  2. #2
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    Re: More Urns and Balls

    The probability of pulling a ball for the first ball (any ball) is 1. The probability of the two balls being the same color is the probability of pulling red, then red, plus white then white, plus green then green:

    $P(A) = 1\cdot \dfrac{5}{12}\cdot \dfrac{5}{12} + 1\cdot \dfrac{4}{12}\cdot \dfrac{4}{12} + 1\cdot \dfrac{3}{12}\cdot \dfrac{3}{12} = \dfrac{25}{72}$

    The probability for the first and third balls to be the same color is similar. We have red, any ball, red. We have white, any ball, white. And we have green, any ball, green.

    $P(B) = \dfrac{5}{12}\cdot 1 \cdot \dfrac{5}{12} + \dfrac{4}{12}\cdot 1\cdot \dfrac{4}{12} + \dfrac{3}{12}\cdot 1\cdot \dfrac{3}{12} = \dfrac{25}{72}$

    Finally, the only way for A and B to both occur is if all three balls are the same color:

    $P(A\cap B) = \dfrac{5^3+4^3+3^3}{12^3} = \dfrac{1}{8}$

    For ii), suppose B has occurred. We have one of three scenarios: red, any ball, red; white, any ball, white; green, any ball, green. What is the probability that it is red, red, red; white, white, white; or green, green, green?

    Well, romsek showed you the formula in the previous problem:

    $P(A|B) = \dfrac{P(A\cap B)}{P(B)} = \dfrac{ \dfrac{1}{8} }{ \dfrac{25}{72} } = \dfrac{9}{25} > \dfrac{25}{72} = P(A)$

    So, knowing that B occurred actually increased the chance that A would occur. This is a case where one's intuition does not necessarily match reality. It is similar to the old game show where there are three doors, only one has a prize behind it. You choose a door. Then, the announcer says let's look behind one of the doors you did not pick (and shows that there is nothing there). Then asks, do you want to stick with your initial choice or do you want to change the door you have chosen? There is a $\dfrac{1}{3}$ chance that the door you initially chose has the prize behind it. There is a $\dfrac{1}{2}$ chance that the door you did not choose has the prize behind it. It is actually more probable that you will win if you change your answer. Moral of the story, conditional probabilities are not always intuitive.
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