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Math Help - calculating probability of making the correct choice

  1. #1
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    calculating probability of making the correct choice

    So an annoying question is floating around where four answers are given, asking the probability of picking the correct one, with the following answers provided:
    a) 25%
    b) 50%
    c) 60%
    d) 25%

    I came up with a question to better visualize this and then I tried to solve that instead.

    We have spheres of 3 colours. Red, Green and Blue.
    There are 10 Red, 5 Green and 10 Blue spheres.
    Assuming a colour is chosen at random, and then a sphere is chosen at random, what is the probability of picking the correct colour sphere?

    The way I approach this is, we have a 1/3 chance of picking each colour and then either a 5/25 or 10/25 chance of picking the right sphere.

    This gives me a probability of picking the right sphere 1/3 of the time. That does not make sense to me since there is a clear bias toward picking a Red or Green sphere. In fact, since the probabilities of picking a sphere always add up to one, changing the amount of spheres does not affect my answer at all. This further seems to indicate that my thought process is wrong. What is the correct way to approach this?

    Help appreciated.
    Thanks,
    Chris
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  2. #2
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    Re: calculating probability of making the correct choice

    Quote Originally Posted by ffezz View Post
    So an annoying question is floating around where four answers are given, asking the probability of picking the correct one, with the following answers provided:
    a) 25%
    b) 50%
    c) 60%
    d) 25%

    I came up with a question to better visualize this and then I tried to solve that instead.

    We have spheres of 3 colours. Red, Green and Blue.
    There are 10 Red, 5 Green and 10 Blue spheres.
    Assuming a colour is chosen at random, and then a sphere is chosen at random, what is the probability of picking the correct colour sphere?

    The way I approach this is, we have a 1/3 chance of picking each colour and then either a 5/25 or 10/25 chance of picking the right sphere.

    This gives me a probability of picking the right sphere 1/3 of the time. That does not make sense to me since there is a clear bias toward picking a Red or Green sphere. In fact, since the probabilities of picking a sphere always add up to one, changing the amount of spheres does not affect my answer at all. This further seems to indicate that my thought process is wrong. What is the correct way to approach this?

    Help appreciated.
    Thanks,
    Chris
    simplify it even further. Consider just two colors blue and yellow. As you said we select one of these randomly (1/2, 1/2)

    Let there be N balls all together and Np of them are blue. The probability of selecting a blue ball is thus p, and selecting a yellow (1-p).

    Given your setup the probability of selecting the correct color ball is $1/2p + 1/2(1-p) = 1/2, ~~ \forall p ~\ni~ 0 \leq p \leq 1$

    At the extremes of p you have 50% chance of either making a certainly correct choice or a certainly incorrect choice.

    Your original logic is correct.
    Thanks from ffezz
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  3. #3
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    Re: calculating probability of making the correct choice

    Interesting Probability Sheet.xlsThis is quite interesting. So because we pick a sphere colour at random, our choice of picking a matching color sphere is the probability of each colour regardless of the actual distribution of spheres. If we have 998 red, 1 blue and 1 green, we still expect to pick the right sphere 1/3 times. Is there a theorem/law for this? There surely must be.
    Ps. I just did a 1000 isntance test using excel, 9999998 reds, 1 blue and 1 green and the result concurred with this. Included is the excel sheet.
    Last edited by ffezz; March 12th 2014 at 02:39 PM.
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  4. #4
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    Re: calculating probability of making the correct choice

    Quote Originally Posted by ffezz View Post
    So an annoying question is floating around where four answers are given, asking the probability of picking the correct one, with the following answers provided:
    a) 25%
    b) 50%
    c) 60%
    d) 25%

    I came up with a question to better visualize this and then I tried to solve that instead.

    We have spheres of 3 colours. Red, Green and Blue.
    There are 10 Red, 5 Green and 10 Blue spheres.
    Assuming a colour is chosen at random, and then a sphere is chosen at random, what is the probability of picking the correct colour sphere?

    The way I approach this is, we have a 1/3 chance of picking each colour and then either a 5/25 or 10/25 chance of picking the right sphere.

    This gives me a probability of picking the right sphere 1/3 of the time. That does not make sense to me since there is a clear bias toward picking a Red or Green sphere. In fact, since the probabilities of picking a sphere always add up to one, changing the amount of spheres does not affect my answer at all. This further seems to indicate that my thought process is wrong. What is the correct way to approach this?

    Help appreciated.
    Thanks,
    Chris
    $P(red\ ball) = \dfrac{10}{25} = \dfrac{2}{5}.$

    $P(green\ ball) = \dfrac{5}{25} = \dfrac{1}{5}.$

    $P(blue\ ball) = \dfrac{10}{25} = \dfrac{2}{5}.$

    $P(red) = P(green) = P(blue) = \dfrac{1}{3}.$

    $P(ball\ chosen\ has\ same\ color\ as\ color\ chosen) =$

    $P(red\ ball\ chosen\ and\ red\ chosen) + P(green\ ball\ chosen\ and\ green\ chosen) + P(blue\ ball\ chosen\ and\ blue\ chosen) =$

    $\dfrac{2}{5} * \dfrac{1}{3} + \dfrac{1}{5} * \dfrac{1}{3} + \dfrac{2}{5} * \dfrac{1}{3} = \dfrac{1}{3} * \dfrac{2 + 1 + 2}{5} = \dfrac{1}{3}.$

    This might be less mysterious (if it is mysterious) by thinking about this

    Suppose all the balls are red. When you choose blue or green you will be wrong. When you choose red you will be right. So you will be right one third the time.
    Last edited by JeffM; March 12th 2014 at 02:46 PM. Reason: typos sigh
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  5. #5
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    Re: calculating probability of making the correct choice

    Thats exactly the way I arived at my solution initially Appreciate all the help
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