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urn question part 3

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  • urn question part 3

    as you can tell, I have problems with urn questions dealing with sampling with replacement. THe following question seems easier than the ones that I previously posted, but I decided to take the approach suggested by the answers in the previous posts to the following question, but I am getting answer greater than one. The question is:

    There are 10 balls in a box: 5 white, 3 red and 2 black. We choose three balls at random with replacement. What is the probability that all 3 balls are of different colors?

    a 0.03 b. 0.09 c. 0.18. d. 0.4 e. 0.84

    My solution: (5/10)(3/10)(2/10)*3!, since for example WRB is different in arrangement than BRW. But this answer yields 180/100

    I was able to initially get the answer by doing (5C1)(3C1)(2C1)/10^3 = 30/100 = 0.03 which matches answer a, but this method contradicts what has been previously posted regarding that this is a sampling with replacement problem, so I am a bit confused here.

    Thank you

  • #2
    Originally posted by audia4
    as you can tell, I have problems with urn questions dealing with sampling with replacement. THe following question seems easier than the ones that I previously posted, but I decided to take the approach suggested by the answers in the previous posts to the following question, but I am getting answer greater than one. The question is:

    There are 10 balls in a box: 5 white, 3 red and 2 black. We choose three balls at random with replacement. What is the probability that all 3 balls are of different colors?

    a 0.03 b. 0.09 c. 0.18. d. 0.4 e. 0.84

    My solution: (5/10)(3/10)(2/10)*3!, since for example WRB is different in arrangement than BRW. But this answer yields 180/100

    I was able to initially get the answer by doing (5C1)(3C1)(2C1)/10^3 = 30/100 = 0.03 which matches answer a, but this method contradicts what has been previously posted regarding that this is a sampling with replacement problem, so I am a bit confused here.

    Thank you
    First of all, I don't see a problem with your first reasoning, except that the numbers you multiplied do not equal 180/100. Try and multiply it out with your calculator. You should get 18/100 = 0.18.

    The same reasoning works with the second method - you've correctly chosen 1 out of each color, but you didn't account for picking them in a different order. So, you still have to multiply 0.03 by 3! = 0.03 * 6 = 0.18. The answer should be C.

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    • #3
      out of 10 balls we have 10c3 possible combinations out of which we need the ones with different colors : (5c1)(3c1)(2c1)

      so (5c1)(3c1)(2c1)/10c3 yields 0.25,not 0.18.What did i do wrong,please?

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      • #4
        Originally posted by Tal View Post
        out of 10 balls we have 10c3 possible combinations out of which we need the ones with different colors : (5c1)(3c1)(2c1)

        so (5c1)(3c1)(2c1)/10c3 yields 0.25,not 0.18.What did i do wrong,please?
        This is under the assumption that the balls are selected without replacement.
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        • #5
          Originally posted by sohpmalvin View Post
          This is under the assumption that the balls are selected without replacement.
          yea,i thought so
          Thank you!

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          • #6
            Well, i think can use multinomial formula which goes like this:
            [(3!/1!1!1!).5*.3*.2]/10C3= .18

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            • #7
              As a general rule, if you have anything resembling 10C3 in a problem *with replacement*, you're doing it wrong.

              Comment


              • #8
                ah nice,i learned so many useful tricks and hints on this forum :smiloe:

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