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Probability to extract balls problem

  1. Mar 22, 2015 #1
    My knowledge about this kind of problems is almost zero, so I need some help.

    I know that if we have 20 balls in a urn, 1 to 20, the probability to extract 1 to 5 in a extraction is 5/20=1/4

    But what if we have 4 extractions with the balls extracted put back in the urn? How do I calculate that ?
  2. jcsd
  3. Mar 22, 2015 #2
    The number of ways of choosing a ball is 20C1. Therefore, the probability of getting a ball (say ball 4) is 1/20

    Now you put it back. Say you are trying to find the probability distribution of getting ball 2 from the urn. The probability of success is 1/20 and probability of failure is 19/20.
    Since you put back the ball, in the next trial, the probability of success and failure is same.
    We call such trials as Bernoulli trials.
    So in 4 trials, the probability of getting ball 2 r times is $$P(X=r)=^4C_r(\frac{1}{20})^r(\frac{19}{20})^{(n-r)}$$
  4. Mar 22, 2015 #3
    If you did not understand the expression, then the topic you have to learn is "probability distribution" and "Bernoulli trials". Use a good mathematics book Or use Wikipedia.
  5. Apr 2, 2015 #4


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    No! If you are saying the probability of choosing the balls 1,2,3,4,5 without replacing the balls back in the urn it goes like this:
    P(extracting ball 1) multiplied by P(extracting ball 2) multiplied by P(extracting ball 3).. Etc.
    (1/20)x(1/19)x(1/18)x(1/17)x(1/16) because once you take ball one out there's only 19 balls left in the urn and so on..!

    However if your talking about extracting ball 1 and replacing it and then choosing ball 2 (with ball one back in the urn) it's out of 20 again. And will go like this...
    P(1) -(probability of choosing ball 1)- times P(2) times P(3).. Etc...

    Now that is the the probablility of choosing the balls in a specific order like, ball 1, then ball 2, then ball 3... So to get the probability of choosing them five balls out of the urn in ANY ORDER (1 then 3 then 5 then 2 then 4 PLUS 1,2,3,4,5 PLUS 2,3,1,4,5 and so on..) then we have to multiply but the number of balls 'factorial' so 5! (! Is the factorial sign, it is the same as 5x4x3x2x1)
  6. Apr 2, 2015 #5
    But I wasnt talking about extracting one ball at a time, but 5 balls at one extraction. Lets consider it an electronic random extraction done by a computer that instantly eliminates 5 balls out of 20.
  7. Apr 3, 2015 #6
    20C5 gives you the number of ways of choosing 5 objects from 20 objects
  8. Apr 3, 2015 #7


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    In that case (like AdityaDev said) 20C5 (20 choose 5) is the number of ways you can choose 5 balls out of 20 without caring about the order you choose them in, so
    Ah i see so like AdityaDev said, it would be 1/(20C5)=0.0000645 (3sf)....?
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