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Count the probability of the event A that the first ball is red

  1. May 12, 2013 #1
    1. The problem statement, all variables and given/known data

    Let there be 2 boxes U1 and U2. Box U1 contains 3 red balls and 4 black balls. Box U2 contains 5 red balls and 3 black balls.
    We chose at random from one of the boxes:
    -If it's a red ball from U1 we put it in U2.
    -If it's a black ball from U1 we put 2 black balls in U2.
    -If it's any ball from U2 we place 3 red balls in U1.
    And then we chose at random from the other box.

    1) Count the probability of the event A that the first ball is red.
    2) What is the probability of the event B that the second ball is red.
    3) What is the probability that both balls are red.
    4) What is the probability of choosing a red ball on the first try while knowing that the second one is red.
    5) Are the events A and B independent.



    3. The attempt at a solution

    After making my tree i got the following:

    1)59/112
    2)41/70
    3)37/112
    4)59/112
    5) The events A and B are independent because: [tex]P_{B}(A)=P(A)=\frac{59}{112}[/tex]

    Is that correct guys?
     
  2. jcsd
  3. May 12, 2013 #2

    Ray Vickson

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    Some aspects are unclear.

    You say "-If it's a black ball from U1 we put 2 black balls in U2." Does this mean that we go and find another black ball from U1 and put it plus the original black, in U2? Or, does it mean that we take our black ball from U1 and then go to a store of spare balls and select another black, then add both of these to U2?

    You say "-If it's any ball from U2 we place 3 red balls in U1." Does that mean we go into U2 and get two more red balls, then add all three to U1? Or, does it mean that we go to a store of spare balls, select two red ones an then put all three reds into U1?

    Finally, you say: "And then we chose at random from the other box." Is the 'other box' the opposite of the one we chose from, or is it the opposite of the one we added balls to?
     
  4. May 12, 2013 #3
    The other box is opposite from the one we chose from first and balls that we put in the other box are from the box you chose from first. For example if it's red from U1 we place that red ball in U2.
     
  5. May 12, 2013 #4

    haruspex

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    I confirm your answers to 1), 2) and 3). Pls show your working for 4).
     
  6. May 13, 2013 #5
    We choose a box at random either U1 or U2 each with a probability of 1/2. Then the probability of choosing a red one from U1 is 3/7 and the probability of choosing a red one from U2 is 5/8. Also since the event of choosing the second ball being red doesn't have any effect on the probability of the first one being red therefore we have:

    P(B|A)=(1/2)*(5/8)+(1/2)+(3/7)=59/112=P(A) And therefore the events A and B are independent.
     
  7. May 13, 2013 #6

    haruspex

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    You're misinterpreting conditionality. The information that the second ball is red could alter the probability that the first ball is red. Consider a much simpler scenario: one urn, two balls, one red and one black. If the two balls are drawn without replacement, and you're told the second drawn was red, what does that tell you about the first ball?
    Use the standard formula: P(A&B) = P(A|B)P(B).
     
  8. May 14, 2013 #7
    Yea that tells me it's black and by the way thank you for you help i found out that the two events are dependent.
     
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