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mridrik

Hey satellite, I your need help with probability again: a bag has 5 red marbles, 3 blue marbles, and two green marbles. 6 marbles are to be drawn from the bag, replacing each one after it is drawn. What is the probability that two marbles of each color will be drawn?

  • 2 years ago
  • 2 years ago

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  1. karatechopper
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    ok well first tell me how many marbles do u have all together?

    • 2 years ago
  2. mridrik
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    10

    • 2 years ago
  3. amistre64
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    yeah, satellite would definnantly be able to breeze by on this one

    • 2 years ago
  4. .Sam.
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    @satellite73

    • 2 years ago
  5. mridrik
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    Yes he is very good with probability I've noticed

    • 2 years ago
  6. amistre64
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    rrbbgg 5/10 + 5/10 + 2/10 + 2/10 + 3/10 + 3/10 would be my guess but im sure its wrong

    • 2 years ago
  7. satellite73
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    ok this is with replacement right?

    • 2 years ago
  8. amistre64
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    i was oring, instead of anding ... among other fauxpauxes

    • 2 years ago
  9. satellite73
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    bag has ten marbles and the number of ways to draw six out of ten is \[\dbinom{10}{6}=\frac{10\times 9\times 8\times 7}{4\times 3\times 2}=210\]

    • 2 years ago
  10. satellite73
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    two marbles of each color. ways of choosing two red is \[\dbinom{5}{2}=10\] number was of choosing 2 out of 3 is \[\dbinom{3}{2}=3\] and the number of way of choosing 2 out of 2 is \[\dbinom{2}{2}=1\]

    • 2 years ago
  11. satellite73
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    so your answer is \[\frac{\dbinom{5}{2}\times\dbinom{3}{2}\times \dbinom{2}{2}}{\dbinom{10}{6}}\] \[\frac{10\times 3\times 1}{210}=\frac{1}{7}\]

    • 2 years ago
  12. mridrik
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    That's incorrect

    • 2 years ago
  13. mridrik
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    way too big according to mathcounts

    • 2 years ago
  14. satellite73
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    yes it certainly is isnt it!

    • 2 years ago
  15. mridrik
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    Mathcounts says the answer is 81/1000, so I know that it had to be 90 times something because I already had 1/4 * 9/100 * 1/25 for the probabilities of drawing 2 red, then two blue, then two green, but I wasn't sure what to multiply by because I didnt know how many ways could each of those be moved around, I finally got 90 by doing 6!/(2!*2!*2!) because 6! is ways of 6 different marbles then divide by 8 because of the 3 repeats :). I just wanted to ask you because I thought you might be able to explain it better. From what I gave you, could you tell me what the probability would be if there wasn't any replacement?

    • 2 years ago
  16. satellite73
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    site clogged up one me

    • 2 years ago
  17. mridrik
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    Do you see my last post on your computer?

    • 2 years ago
  18. satellite73
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    yeah i messed up and did the problem as if it was without replacement

    • 2 years ago
  19. mridrik
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    Can you show me how to do it using "my way" if there were no replacements?

    • 2 years ago
  20. satellite73
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    ok sorry it took a while i had to think

    • 2 years ago
  21. satellite73
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    here we go. rr gg bb has probability \[\frac{5}{10}\times \frac{5}{10}\times\frac{3}{10}\times \frac{3}{10}\times \frac{2}{10}\times \frac{2}{10}=\frac{3^3}{1000}\]

    • 2 years ago
  22. satellite73
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    then we have to figure out the possible combinations of 2 r, 2 g and 2 b if we could tell them all apart there would be 6! possibilitites, but since we cannot distinguish between the two r, the two g and the 2 b there are \[\frac{6!}{2!\times 2!\times 2!}=\frac{6\times 5\times 4\times 3\times 2}{2\times 2\times 2}\] \[=3\times 5\times 2\times 3\] different ways to rearrange this

    • 2 years ago
  23. satellite73
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    so our final answer is \[\frac{3^2\times 5\times 2\times 3^2}{10000}\] \[=\frac{81}{1000}\]

    • 2 years ago
  24. satellite73
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    if there are no replacements the answer i wrote above would work (way above) to do it "your way" we would compute as follows

    • 2 years ago
  25. satellite73
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    \[\frac{5\times 4\times 3\times 2\times 2\times 1}{10\times 9\times 8\times 7\times 6\times 5}\times 3\times 5\times 2\times 3\] the first because that is the probability you ge t2r 2g 2b in that order (without replacement) and the second term because that is the number of ways you can arrange rr gg bb as before the answer \[\frac{1}{7}\]as with my first post as you can verify

    • 2 years ago
  26. satellite73
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    sorry i messed up the first time, i didn't read it correctly

    • 2 years ago
  27. mridrik
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    It's all right, but can you explain your last post a little bit better please

    • 2 years ago
  28. satellite73
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    i can try

    • 2 years ago
  29. satellite73
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    first method i think is easier, the one where i wrote \[\frac{\dbinom{5}{2}\times\dbinom{3}{2}\times \dbinom{2}{2}}{\dbinom{10}{6}}=\frac{10\times 3\times 1}{210}=\frac{1}{7}\]

    • 2 years ago
  30. mridrik
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    I don't really understand that one either :(

    • 2 years ago
  31. satellite73
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    ok that one is easy to explain

    • 2 years ago
  32. satellite73
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    you are picking six out of ten and the number of ways to pick six out of ten is by definiton \[\dbinom{10}{6}\] which you calculate by \[\frac{10\times 9\times 8\times 7}{4\times 3\times 2}=210\]so that is our denominator. you might notice that i computed \[\dbinom{10}{4}\] instead of \[\dbinom{10}{6}\] because they are obviously the same and it is easier so that is your denominator, all the possible ways to choose 6 out of 10

    • 2 years ago
  33. satellite73
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    now you have 5 red and you are choosing 2 out of that 5 and the number of ways you can do that is \[\dbinom{5}{2}=\frac{5\times 4}{2}=10\] and you are choosing 2 out of 3 blue, number of way you can do that is clearly 2 and only one way to choose 2 out of 2 green so you you get the numerator of \[10\times 3\times 1=30\] 30 ways out of a total of 210 give the probability of \[\frac{30}{210}=\frac{1}{7}\]

    • 2 years ago
  34. mridrik
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    Okay I see where you get the numbers but I don't see how that gives you the probability

    • 2 years ago
  35. satellite73
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    your method is to do the following we compute the probability you get exactly this sequence of red, blue, and green r r b b g g

    • 2 years ago
  36. satellite73
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    oh that is the "counting principle" number of way to pick two out of 5 is 10, number of ways to pick 2 out of 3 is 3, number of ways to pick 2 out of 2 is 1, by the counting principle the number of ways we can do this together is 10*3*1=30 take that out of the total number of ways you can get 6 out of 10 gives 30/210

    • 2 years ago
  37. satellite73
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    now back to your method. we compute the probability we get exactly the sequence rr bb gg

    • 2 years ago
  38. mridrik
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    right I understand that but I dont understand how that gives you the probability that 2 of each color is drawn w/o replacement

    • 2 years ago
  39. mridrik
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    Here's an example of the way I see it how you explain your way: ...to find that probability you take the colors and you make a rainbow, since rainbows have 7 colors, the answer is 1/7.

    • 2 years ago
  40. satellite73
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    first is red is \[\frac{5}{10}\] second is red given first is red is \[\frac{4}{9}\] probability that third is blue given first two are red is \[\frac{3}{8}\] probability that the fourth is blue given first is red, second is red third is blue is \[\frac{2}{7} \] probability that fifth is green given all that other stuff is \[\frac{2}{6}\] probability that sixth is green given the above is \[\frac{1}{5}\]

    • 2 years ago
  41. satellite73
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    multiply these all together and then multiply by the number of arrangement of r r b b g g and you get the same answer

    • 2 years ago
  42. satellite73
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    lets back up for a second total number of ways you can pick 6 out of 10 is 210 so there are that many elements in your sample space and that is your denominator since they are equally likely

    • 2 years ago
  43. satellite73
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    then by the counting principle there are 10*3*1 = 30 ways to pick 2 red,2 green and 2 blue so that is your numerator that simple

    • 2 years ago
  44. mridrik
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    wait, hold on stop for a moment, un momento!,

    • 2 years ago
  45. satellite73
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    ok

    • 2 years ago
  46. mridrik
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    so your 5th to last post (starts off with "first is red is") that would also give me 1/7?! Multiplying all of that by 90? My goodness it does! Okay, now back to explaining your way please.

    • 2 years ago
  47. satellite73
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    ok first is red there are 5 red, 10 total, so probability first is red is 1/2 yes?

    • 2 years ago
  48. mridrik
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    yes.

    • 2 years ago
  49. satellite73
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    then we compute the probability that second is red GIVEN that first is red. that is 4/9 right?

    • 2 years ago
  50. satellite73
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    so probability that first is red AND second is red is \[\frac{5}{10}\times \frac{4}{9}\]

    • 2 years ago
  51. mridrik
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    1/2 * 4/9, yes

    • 2 years ago
  52. satellite73
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    ok now third is blue given first two are red. 3 blues, 8 left, so \[\frac{3}{8}\]

    • 2 years ago
  53. satellite73
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    likewise fourth this blue given first red, second red, third blue is \[\frac{2}{7}\]

    • 2 years ago
  54. mridrik
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    yes

    • 2 years ago
  55. satellite73
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    fifth green given first red, second red, third blue, fourth blue \[\frac{2}{6}\]

    • 2 years ago
  56. mridrik
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    then 1/5 i gotcha

    • 2 years ago
  57. satellite73
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    and finally sixth is green given blah blah is \[\frac{1}{5}\] right

    • 2 years ago
  58. satellite73
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    now we compute the probabilty that all this occurs by multiplyiing it all together

    • 2 years ago
  59. satellite73
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    \[\frac{5}{10}\times \frac{4}{9}\times \frac{3}{8}\times\frac{2}{7}\times \frac{2}{6}\times \frac{1}{5}\] gives the probability we get R R B B G G in that order

    • 2 years ago
  60. satellite73
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    but order does not matter, we just want 2 red, 2 blue and 2 green

    • 2 years ago
  61. satellite73
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    so now the question is, how many different arrangments of r r b b g g are there? and the answer is \[\frac{6!}{2\times 2\times 2}=\frac{6\times 5\times 4\times 3\times 2}{2\times 2\times 2}\] \[=6\times 5\times 3\]

    • 2 years ago
  62. mridrik
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    so multiply by 90 right. 6!/2!^3

    • 2 years ago
  63. satellite73
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    i get 90, yes

    • 2 years ago
  64. satellite73
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    each combination has same probability, (you can check this) so you multiply the number above by 90

    • 2 years ago
  65. mridrik
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    it's interesting how you can switch around each of the marbles such as b g r r g b which is 3/10 * 2/9 * 5/8 * 4/7 * 1/6 * 2/5 * 90 and still get the same answer, i think that is so cool

    • 2 years ago
  66. satellite73
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    yes they give the same thing it is true!

    • 2 years ago
  67. satellite73
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    so take \[\frac{5}{10}\times \frac{4}{9}\times \frac{3}{8}\times\frac{2}{7}\times \frac{2}{6}\times \frac{1}{5}\] multipliy by 90 and get \[\frac{1}{7}\]

    • 2 years ago
  68. mridrik
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    So now I still really do not understand your way, i know how you got the numbers, but, like my analogy i made i just dont see how you put it together and get 1/7

    • 2 years ago
  69. satellite73
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    ok i don't want to annoy you with it but the counting principle is easy

    • 2 years ago
  70. satellite73
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    if there are 10 ways to do one thing and 3 ways to do another, then there are 30 ways to do them together

    • 2 years ago
  71. satellite73
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    so there are 30 ways to select 2 reds and 2 blues

    • 2 years ago
  72. mridrik
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    I understand the counting principle fine, what I do not understand is how you used it to come up with your answer

    • 2 years ago
  73. satellite73
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    since there is only one way to get 2 out of 2 greens there are still 30 ways to get 2 reds, 2 blues and 2 greens

    • 2 years ago
  74. satellite73
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    so 30 ways to perform your task and 210 possible ways to choose 6 out of 10

    • 2 years ago
  75. satellite73
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    since they are presumed to be equally likely probability is 30/210

    • 2 years ago
  76. satellite73
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    counting prinicple tells you the number of ways you can do something. if they are equally likely then probabilty is simple enough

    • 2 years ago
  77. mridrik
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    Well you seem to be pretty efficient with your way but alas, I'll just stick with the way you showed me how to do my way lol. It makes more sense inside my head

    • 2 years ago
  78. mridrik
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    Thanks for your help again satellite, I'll try to find this one probability problem I've been looking for that is the hardest I've ever seen

    • 2 years ago
  79. satellite73
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    ok it works, although it is faily annoying to have to compute \[\frac{6!}{2^3}\] when it is not necessary

    • 2 years ago
  80. satellite73
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    yw

    • 2 years ago
  81. mridrik
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    i has calculator that does it for me, plus I know the formula

    • 2 years ago
  82. satellite73
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    lol

    • 2 years ago
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