1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Conditional Probability and Urns in balls

  1. Feb 2, 2012 #1
    1. The problem statement, all variables and given/known data
    You have 3 urns: Urn 1 has 3 red balls, Urn 2 has 2 red balls, 1 blue ball. Urn 3 has 2 red balls 2 blue balls. You pick a ball from each urn and place it into Urn 4.
    You draw 2 balls from Urn 4 and they are red. What is the conditional probability that the 3rd ball is also red?

    2. Relevant equations
    Bayes law?


    3. The attempt at a solution
    The third ball can be from urn 1, 2, or 3. So I initially tried
    P(3rd red given first 2 are red)=P(3rd ball is red and from Urn 1)+P(3rd ball is red and from Urn 2)+P(3rd ball is red and from Urn 3)
    =1/3(1+2/3+1/2)=13/18.

    But this doesn't seem right, I feel like I'm neglecting the conditions of the first 2.
    what if I try P(3rd red given first 2 are red)=P(1st two balls being red from urn 1 and urn 2)*P(3rd ball red from urn 3)+P(1st two balls being red from urn 1 and urn 3)*P(3rd ball red from urn 2)+ P(1st two balls being red from urn 2 and urn 3)*P(3rd ball red from urn 1)

    Is that formulation correct?
     
  2. jcsd
  3. Feb 2, 2012 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    By far the easiest (and most error-free) way is to get the probabilities of the various numbers of red in Urn 4, and to forget about which of Urns 1-3 they came from. Letting R = number red in Urn 4, we have R = 1, 2 or 3: we always have at least 1 red in U4 because all balls in U1 are red. {R=1} = {blue from U2 & blue from U3}, {R=2} = {red from U1 and blue from U2} or {blue from U1 & red from U2}, and {R=3} = {red from U2 & red from U3}. Once you know the values of P{R=1}, P{R=2} and P{R=3}, you can find P{R=3 | R >= 2}, which is what you want.

    RGV
     
  4. Feb 2, 2012 #3
    Let's see:
    P(R=1)=1*1/3*1/2=1/6
    P(R=2)=1*2/3*1/2+1*1/3*1/2=1/2
    P(R=3)=1*2/3*1/2=1/3
    which add to 1 as desired.

    Then, by Bayes' Rule: P(R=3|R>=2)= P(R=3)/(P(R=2)+P(R=3))=2/5

    ---My first question is: did I use Bayes rule correctly? It seems, correctly, that P(R=1) is irrelevant given the condition.

    I checked this answer a different way: we have 3*3*4=36 possible outcomes of 3 balls.
    Of these,3*2*2=12 have 3 red balls, 3*2*2+3*1*2=18 have 2red, 1 white, and 6 have 1 red 2 white.

    Restricting ourselves to >=2 reds, we have 12/(12+18)=2/5, which agrees.

    ---My 2nd question is: we are findingP{R=3 | R >= 2} the probability that 3 balls are red, given at least 2 are. Is this the same as finding the probability that the 3rd ball, given 1st are 2 red?

    Thanks
     
  5. Feb 2, 2012 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    OK, I see what you mean: the events {R >= 2} and {1st two red} could be different, because one of the possibilities in {R >= 2} is RBR. So *my second suggestion was incorrect*. Instead, compute P{1st red and 2nd red} as P{1st red and 2nd red|R=2}*P{R=2} + P{1st red & 2nd red | R=3}*P{R=3}. We know P{R=2} and P{R=3}. We have P{1st red & 2nd red | R=3} = 1 of course, and getting P{1st red & 2nd red | R=2} is not very hard. Note that here, 1st and 2nd refer to drawings from U4, not from U1--U3.

    RGV
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Conditional Probability and Urns in balls
Loading...