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Math Help - Probability of selecting the first urn.

  1. #1
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    Question Probability of selecting the first urn.

    Ok here it goes,
    One urn contains 3 red and 7 white marbles. A second urn contains 2 red and 8 white marbles. One urn is selected and a ball drawn. The probability of selecting the first urn is 0.4 and the probability of selecting the second is 0.6. If a white ball is drawn, find the probability it came from the second urn.

    3 red 7 white, urn 1= 10
    2 red 8 white, urn 2= 10
    P(urn_1)=0.4, P(urn_2)=0.6 0.4*0.6=24
    looks like to me urn 2 has a better chance than 1 becaue of th 60%, both have 10 marbles but urn 2 has one more white one than urn 1. List of possible answers are -

    1. 0.800
    2. 0.632
    3. 0.560
    4. 0.480 From here I'm stuck in the mud. Need help with the rest of it. thanks for looking over my problem.
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  2. #2
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    Smile man that was easy

    I tryed everything I could to come up with an answer, I didn't think about that % is all that was needed. Thanks soooo much for showing me that. I have a few more like this I have to do, they all will work out like that. Thanks again Galactus for all your help.
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  3. #3
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    Hello, kwtolley!

    I hope you're familiar with Bayes' Theorem . . .


    Urn A contains 3 red and 7 white marbles.
    Urn B contains 2 red and 8 white marbles.
    One urn is selected and a ball drawn.
    The probability of selecting urn A is 0.4
    and the probability of selecting urn B is 0.6.

    If a white ball is drawn, find the probability it came from the urn B.

    1)\;0.800\qquad2)\;0.632\qquad3)\;0.560\qquad4)\;0  .480

    Bayes' Theorem: . P(\text{urn B }|\text{ white}) \;= \;\frac{P(\text{urn B }\land\text{ white})} {P(\text{white})}

    The numerator is:
    . . P(\text{urn B }\land\text{ white})\:=\:P(\text{urn B})\cdot P(\text{white}) 0.6)(0.8)\:=\:0.48" alt=" \:=\0.6)(0.8)\:=\:0.48" />

    The denominator is:
    . . P(\text{white}) \:=\:P(\text{urn A }\land\text{ white}) + P(\text{urn B }\land\text{ white})
    . . . . . . . . . =\qquad\quad(0.4)(0.7) \qquad+ \qquad(0.6)(0.8)\quad =\quad0.28  + 0.48 \;= \;0.76


    Therefore: . P(\text{urn B }|\text{white}) \;= \;\frac{0.48}{0.76} \;= \;\frac{12}{19} \;=\;0.631578947

    Answer (2): . 0.632

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  4. #4
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    Soroban, I sure had a brain fart

    I even ran Bayes theorem and thought "naaaaa". I checked back to fix my stupidity and you had taken care of it. Good!. Sorry kwtolley
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  5. #5
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    Smile thanks and yes

    Yes I have read Bayes Theorem but I need to see it on paper how to set up a problem the first time. Now I see it and I have some more to do. I can say one thing it's not as easy as it looks if you ask me.
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