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If there is a box which has 7 red balls

  1. May 5, 2005 #1
    If there is a box which has 7 red balls, 4 white balls and 9 black ball and you take 3 balls out what is the probability of the three balls being red and the probability of the balls being 2 white and 1 red
     
  2. jcsd
  3. May 5, 2005 #2

    honestrosewater

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    What is your sample space? What is the probability of picking one red ball? If you don't return that red ball to the box, what is your new sample space? What is the probability of picking another red ball? ...
     
  4. May 5, 2005 #3

    HallsofIvy

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    I assume you are talking about "sampling without replacement"- that is you do NOT put a ball back into the box after noting its color.

    1. In order to get "all three ball being red", the first ball must be red. There are a total of 7+ 4+ 9= 20 ball, 7 of them red so the probability of that is 7/20.
    Once that has happened, there are 19 balls left, 6 of them red. The probability that the second ball you pick will be red is 6/19. The probability that the first two balls are both red is the product: (7/20)(6/19).
    If you have done that, then there are 18 balls left, 5 of them red: the probability that the next ball is also red is 5/18. The probability of picking 3 red balls is
    (7/20)(6/19)(5/18) which can be reduced (since 5/20= 1/4 and 6/18= 1/3) to
    (7/4)(1/19)(1/3)= 7/228 or about 0.0307.

    "2 white and one red" is just a little harder. One way for that to happen is to get "white, white, red" in that order. Since there are 4 white balls, the probability that the first ball will be white is 4/20= 1/5. If the first ball is white, then there are 19 balls left and 3 of them are white. The probability that the second ball is also white is 3/19. Assuming that the first two balls were white, there are now 18 balls left but 7 of them are still red. The probability that the third ball will be red is 7/18. The probability of gettin "white,white,red", in that order, is (1/5)(3/19)(7/18).

    But drawing "white, red, white" would also give "2 white and 1 red". As before the probability that the first ball is white is 4/20= 1/5. Assuming the first ball is white then on the second draw there are 19 balls left and 7 of them are red. The probability that the second ball is red is 7/19. Finally, if the draws have happened that way there are 18 balls left, of which 3 are still white. The probability that the third ball is white is 3/18= 1/6. The probability of getting "white, red, white" in that order is (1/5)(7/19)(3/18). Notice that the denominators are exactly the same as in the previous case! Of course: that was always the total number of balls and it didn't matter what color was taken out. Even more, the numerators are the same- just their order is changed. That's because we are taking out the same balls- just changing the order. But multiplication is "commutative"- order doesn't matter. The point is that the probability of getting these balls in any specific order is the same: (1/5)(3/19)(7/18). To find the probability of getting "white, white, red" is any order, we just have to recognize that there are 3 "orders": (white, white, red), (white, red, white), and (red, white, white) so the probability of getting 2 white and 1 red ball, in any order is 3(1/5)(3/19)(7/18)= (1/5)(1/19)(7/2)= 7/190 or about 0.0368.
     
  5. May 5, 2005 #4
    thanx that really helped
     
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