Probability

Discussion in 'Math' started by Biggsy100, Aug 11, 2014.

  1. Biggsy100

    Thread Starter Member

    Apr 7, 2014
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    A bag containing 1 green, one red and 5 white balls.

    I withdraw a ball and then without replacing it withdraw another. What is the probability that I withdraw both red and green balls?

    So, I know:
    Red = 1/7
    Green = 1/7
    White 4/7 (?)

    For the sake of the equation, do I ignore the white balls, thus give it as:

    1/7 * 1/7 = 1/49 or would the 2 chances merge to make 2/49?
     
  2. NorthGuy

    Active Member

    Jun 28, 2014
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    Calculate the total number of possible two-ball combinations - 21. 1/21 then.
     
  3. joeyd999

    AAC Fanatic!

    Jun 6, 2011
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    The chance of drawing a red on the first try is 1/7, yes?

    Also, the chance of drawing green on the first try is 1/7, yes?

    So, the chance of drawing red -or- green on the first try is what?

    Then, 6 balls remain, with either a single red or a single green.

    What are the odds of drawing the remaining colored ball?

    Combine the first and second answer (i.e. multiply).
     
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  4. Aerb

    New Member

    Jul 9, 2014
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    i think you're confused and you made me confused as well with your last comment so i'll just do the math straight forward with explanations. Here we go;

    If i get you right, we have 2 desired situations here. I'll handle them separately.
    Situation 1;
    You withdraw a red ball first with the probability of 1/7. Then you repeat it without dropping it back. Now you withdraw a green ball with the probability of 1/6. So, for situation 1, your desired probability is (1/7)*(1/6)=1/42
    Situation 2;
    You withdraw a green ball first with the probability of 1/7. Then you repeat it without dropping it back. Now you withdraw a red ball with the probability of 1/6. So, for situation 2, your desired probability is (1/7)*(1/6)=1/42

    The last step is summing these probabilities up, since there is a 'or' relation between situation 1 and situation 2;
    (1/42)+(1/42)=1/21

    Hope this helps. If you need any further explanation, feel free to ask.
     
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  5. THE_RB

    AAC Fanatic!

    Feb 11, 2008
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    What math did you use to get 21? Just curious. I used Aerb's solution which I thought was the quickest way to solve.
     
  6. Biggsy100

    Thread Starter Member

    Apr 7, 2014
    88
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    This is 'one of them' ones that I understand the principle but have a git of a time trying to put on paper - Thanks.

    So I have 2 more senarios;

    1 -I replace the balls, shake the bag and then withdraw 2 balls together. What is the probability that I have with drawn both the red and green balls?


    2 - All the balls are returned to the bag and 4 are removed together. What si the probability that both red and green balls are left in the bag?

    For both of these, is it a simple case of using 2 and respectively instead of 1?
     
  7. djsfantasi

    AAC Fanatic!

    Apr 11, 2010
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    Use the "Binomial Coefficient" to calculate the number of combinations of r objects of n items, where r=2 and n=7.

    [​IMG]

    \frac{7!}{2!(7-2)!} = \frac{5040}{2*120}

    This is equal to 21!
     
    Last edited: Aug 12, 2014
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  8. djsfantasi

    AAC Fanatic!

    Apr 11, 2010
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    For your latest scenarios, the first is identical to the original problem, since no matter what, the second ball (even if picked at the same time) is still coming from a set of 6 elements.

    The second scenario can be thought of as the inverse of the first.
    Hence if P_{1} is equal to the probability of the first scenario, then 1-P_{1} is the probability of the second.
     
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  9. NorthGuy

    Active Member

    Jun 28, 2014
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    That's the formula I used. But it's simpler than it seems because n!/(n-r)! is quite easy to calculate. If r is 1, it's n. If r is 2, it's n*(n-1). If r is 3, it's n*(n-1)*(n-2) etc.
     
  10. NorthGuy

    Active Member

    Jun 28, 2014
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    What is the number of possible combinations of 4 balls out of 7 - 35.

    How many meet our criteria? It is equal to the number of combinations of 4 out of 5 white balls - 5.

    Probability: 5/35 = 1/7.

    Just to make sure you didn't make any mistakes: you'll get 10 situations where you get a green ball with all others white, 10 situations where you get a red ball and all others white, and 10 situations where you get green, red and two whites.

    10+10+10+5=35.

    Do they make you draw the real balls many many times and verify that the result converges to expected probabilities?
     
  11. Biggsy100

    Thread Starter Member

    Apr 7, 2014
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    Just to confirm, you mean this: All the balls are returned to the bag and 4 are removed together. What is the probability that both the red and green balls are left in he bag?

    - This you say is the same method as the first?
     
  12. Biggsy100

    Thread Starter Member

    Apr 7, 2014
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    Please can you clarify how 5/35 = 1/7 is there an equation ?
     
  13. NorthGuy

    Active Member

    Jun 28, 2014
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    No, there's a table. You had to memorise it at school: 5 x 7 = 35. I don't know if they still teach it ...
     
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  14. joeyd999

    AAC Fanatic!

    Jun 6, 2011
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    I was going to reply in a similar fashion, but it just seemed harsh. Even for me.
     
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  15. Biggsy100

    Thread Starter Member

    Apr 7, 2014
    88
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    Us Brits say you North Americans have no sense of humour.....

    Regardless, coming from educated folk such as yourself still **** poor. My Bad....:p
     
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  16. NorthGuy

    Active Member

    Jun 28, 2014
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    Yes, English sense of humor in times of Charles Dickens was so good that some Brits still believe they have it :)
     
  17. Biggsy100

    Thread Starter Member

    Apr 7, 2014
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    Ha...which is it English or British? Do you know the difference? ;)
     
  18. NorthGuy

    Active Member

    Jun 28, 2014
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    Who? Famous sense of humor? Definitely English. Dickens? Both English and British at the same time. Brits? Some are still English ... :)
     
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  19. Biggsy100

    Thread Starter Member

    Apr 7, 2014
    88
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  20. NorthGuy

    Active Member

    Jun 28, 2014
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    It'll be more if Scottland separates.

    Real-life application of probabilities :)
     
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