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Confused about basic probability

  1. Nov 12, 2009 #1
    A box contains 5 white balls, 3 red balls and 2 green balls.

    The balls are taken from box. With replacement and taking a total of 3 balls,

    P(all balls have different colour)=(5/10)x(3/10)x(2/10) x 3! =0.18

    This answer is the same as at the back of the book. My explanation is that a white, red or green ball must be taken and there are 3! ways of arranging the 3 scenarios.

    Is this correct????

    Then secondly,
    Without replacement and taking a total of 5 balls

    My wrong method:
    P(exactly 2 balls are white) = (5/10)x(4/9)x(5/8)x(4/7)x(3/6) x 5!

    This probability cannot be correct since it is above 1. What is the loophole that I did not see???

    I dont need the answer to this question (I can solve using counting total number of cases) but I want to know the loophole in my thinking.
     
    Last edited: Nov 12, 2009
  2. jcsd
  3. Nov 12, 2009 #2

    Gib Z

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    I think your first question isn't correct. It says WITHOUT replacement, so the chance of a white ball is 5/10, then a red is 3/NINE, and green is 2/EIGHT, then you multiply by 3!.

    For the next one, is that 't' meant to be a variable, or a typo for "two". You should give logical justifications as to where the numbers in your solution come from, it's not always obvious.
     
  4. Nov 12, 2009 #3
    Opps. The first part is WITH REPLACEMENT. My bad. Apologies.

    And t is a typo. Its supposed to be 5.

    Since possible ways of taking 5 balls with 2 without replacement is

    (5/10) (4/9) [Assume these are white balls] (5/8)(4/7)(3/6) [Remaining picks for colours besides white] x5! [Since the anyone scenario can occur in any order.

    But I get an answer above 1! What is the loophole in my thinking?
     
    Last edited: Nov 12, 2009
  5. Nov 12, 2009 #4

    Gib Z

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    You account for changing arrangements of the picks by multiplying by 5!, but you didn't account for repetitions.
     
  6. Nov 12, 2009 #5
    Thank you.

    But Im still quite confused about repetitions. Can you give an example to 'see' the repetition?

    Also, wouldnt there be repetition for the first part which got correct?

    Thanks.
     
  7. Nov 12, 2009 #6

    Gib Z

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    For the first one, all the balls are different colors and so distinct from each other, so there is no repetition. For the next question, the two white balls picked are identical - the order they are placed in does not matter. Also, the other 3 balls, depending on the cases, will also have this sort of repetition.
     
  8. Nov 12, 2009 #7
    Thanks for your clarification.

    I think I can see it much better now but I still wonder when to put factorial signs behind probability (because sometimes I see some question's answer have and some don't).

    Can you provide me with greater insight about this? When is the arrangement important and when it is not?

    Thank you so much.
     
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