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Probability problem

  1. Sep 28, 2011 #1
    1. The problem statement, all variables and given/known data
    There are 3 white and 2 black balls in box. We pick up 2 balls from the
    box without replacement and one of them appears to be white. Find probability that the
    second ball is also white.


    2. Relevant equations



    3. The attempt at a solution
    can you help me with this problem. i know it may not be too difficult, but im a beginner and got stucked a bit.
    thanx
     
  2. jcsd
  3. Sep 28, 2011 #2

    NascentOxygen

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    Suggestion: put 5 balls, labelled W and B, in a box. Shake, then select two. If one is white, add a tick on a sheet of paper where you are keeping score. If the second one is also white, circle that tick. Repeat 50 more times. What proportion of your ticks are circled?

    Repeat once more, 50 times.

    Compare the result with the theoretical value, when you find out how to calculate it.
     
  4. Sep 28, 2011 #3

    HallsofIvy

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    "Selecting two balls at once" is exactly the same as selecting the balls one at a time.

    If you had 3 white and 2 black balls in a box, and took one of the white balls out, you would have left 2 white and 2 black balls. If you now take one of the remaining balls out, what is the probability it will be white?
     
  5. Sep 28, 2011 #4
    P(A│B)= (P(A∩B))/(P(B)) - formula for conditional probability

    the result i got is P(A│B)= (P(A∩B))/(P(B))=27/10 is it correct?
     
  6. Sep 28, 2011 #5
    27/10 = 2.7. Probabilities must be between 0 and 1 (inclusive). This should tell you right away you did something wrong.
     
  7. Sep 28, 2011 #6
    thanx a lot for pointing out this, so p(a/b)=3/10:9/10=3/10*10/9=3/9=1/3
     
  8. Sep 29, 2011 #7

    NascentOxygen

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    Given: you have a ball in your hand, and it's white.
    Find: the probability that a ball chosen from those others is also white.

    In choosing that second ball, there are 2 white balls and 2 black. P(w) = 2/4.
     
  9. Sep 29, 2011 #8
    This statement can be interpreted in at least two ways.

    (1) I picked up two balls and the first one I looked at was white.

    (2) I picked up two balls and at least one was white.

    It makes a difference. I'm not sure, but I lean toward (2) as the intended interpretation, mostly because this is a problem about conditional probability and authors love to throw in that kind of curve ball.
     
  10. Sep 30, 2011 #9

    HallsofIvy

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    No, it doesn't make any difference at all. That was what I said in post # 3. If a box contains 2 black and 3 white balls, and you take one of the white balls out, you are left with 2 black and 2 white balls in the box. If take out one of them, at random, what is the probability that it will be white? That is exactly what this question is asking.
     
  11. Sep 30, 2011 #10

    LCKurtz

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    I agree with Awkward on this. I think the interpretation of the question is not clear, possibly because the OP didn't word it carefully, and the two interpretations are not equivalent. It's the same distinction as in the usual problem

    1. A family has two children. The oldest is a boy. What is the probability that the youngest is a boy?

    2. A family has two children. One of them is a boy. What is the probability that the other is? And, of course, the answers are different.
     
  12. Sep 30, 2011 #11

    HallsofIvy

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    Now, I agree that those are different. But that is not the same as the previous question.
     
  13. Sep 30, 2011 #12
    Not to be rude but I think this question is being made out to be much more complicated by some of the posters than it actually is.

    If you start off with balls W W W B B and 1 white ball is picked out, you are left with W W B B. The question is asking you, if you have W W B B, what is the probability that you pick out a white ball.
     
  14. Sep 30, 2011 #13

    HallsofIvy

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    Yes, I've said that twice now!
     
  15. Oct 1, 2011 #14

    LCKurtz

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    Sorry, I'm not going to concede that. It does not say you picked one and examined it and found it to be white. It says you pick two balls and one of them is white. It does not say exactly one of them is white, which would make the question trivial anyway, so a reasonable interpretation is "at least one is white" or "there is a white one".
     
  16. Oct 1, 2011 #15

    LCKurtz

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    And I should have added, with this interpretation of the problem, the answer is 1/7, not 1/2.
     
  17. Oct 1, 2011 #16

    LCKurtz

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    I can't tell from your two posts whether you expect the results of the experiment will verify the P = 1/2 result or whether you expect the OP to discover for himself that they won't agree. They won't, of course, because the experiments are not the same. And the first experiment will verify the 1/7 answer.
     
    Last edited: Oct 1, 2011
  18. Oct 10, 2011 #17

    NascentOxygen

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    Mind explaining the 1/7 answer?

    I do expect the results to be identical since I am describing the same procedure but in different words. It is an easy experiment to carry out, and it would be a good one to set for a class. To test it out, I performed it 3 times, making 50 draws each.

    Test 1: 3 draws brought up only black balls, so these draws don't count. Of the 47 draws which brought up at least one white ball, 12 involved a pair of white balls.
    Test 2: 4 BB draws disqualified. Of the 46 remaining, 21 involved a pair of white balls.
    Test 3: 3 BB draws disqualified. Of the 47 remaining, 30 involved a pair of white balls.

    Grand total: 63/140

    There really is no point in keeping a count of the BB events, other than to illustrate just how rare an event it is that a draw brings up no whites.
     
  19. Oct 10, 2011 #18

    LCKurtz

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    The experiment is to draw two balls. Let A be the event a white ball is drawn and B be the event that both balls are white.

    There are C(5,3) = 10 ways to draw two balls, and C(3,2) = 3 ways to draw two black, leaving 7 ways to get a white ball.

    [tex]P(B|A) = \frac {P(B\cap A)}{P(A)}= \frac{\frac 1 {10}}{\frac 7 {10}}=\frac 1 7[/tex]
     
    Last edited: Oct 10, 2011
  20. Oct 11, 2011 #19

    NascentOxygen

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    How should I change the procedure so the experiment will support the 1/7 figure?
     
  21. Oct 11, 2011 #20

    LCKurtz

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    Looking at again, I'm afraid I did the calculation for 3 black and 2 white instead of 3 white and 2 black. The explanation should read:

    There are C(5,2) = 10 ways to draw two balls, C(2,2) = 1 way to draw two black, leaving 9 ways to get a white ball. There are C(3,2) = 3 ways to get two white balls.

    [tex]P(B|A) = \frac {P(B\cap A)}{P(A)}= \frac{\frac 3 {10}}{\frac 9 {10}}=\frac 1 3[/tex]

    You can also look at it by cutting the sample space down to the 9 pairs that have a white ball and observe that 3 of them have two whites.
    I'm guessing your experiment just needs more trials to approach 1/3. Sorry about the 1/7 error.
     
    Last edited: Oct 11, 2011
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