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mridrik

  • 2 years ago

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?

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

  2. mridrik
    • 2 years ago
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    10

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

  4. .Sam.
    • 2 years ago
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    @satellite73

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

  6. amistre64
    • 2 years ago
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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

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

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

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

  10. satellite73
    • 2 years ago
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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 \[\dbinom{2}{2}=1\]

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

  12. mridrik
    • 2 years ago
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    That's incorrect

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

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

  15. mridrik
    • 2 years ago
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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?

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

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

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

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

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

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

  22. satellite73
    • 2 years ago
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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

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

  24. satellite73
    • 2 years ago
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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

  25. satellite73
    • 2 years ago
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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

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

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

  28. satellite73
    • 2 years ago
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    i can try

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

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

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

  32. satellite73
    • 2 years ago
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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

  33. satellite73
    • 2 years ago
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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 \[\frac{30}{210}=\frac{1}{7}\]

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

  35. satellite73
    • 2 years ago
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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

  36. satellite73
    • 2 years ago
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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

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

  38. mridrik
    • 2 years ago
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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

  39. mridrik
    • 2 years ago
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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.

  40. satellite73
    • 2 years ago
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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 \[\frac{1}{5}\]

  41. satellite73
    • 2 years ago
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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

  42. satellite73
    • 2 years ago
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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

  43. satellite73
    • 2 years ago
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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

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

  45. satellite73
    • 2 years ago
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    ok

  46. mridrik
    • 2 years ago
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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.

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

  48. mridrik
    • 2 years ago
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    yes.

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

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

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

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

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

  54. mridrik
    • 2 years ago
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    yes

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

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

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

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

  59. satellite73
    • 2 years ago
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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

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

  61. satellite73
    • 2 years ago
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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 \[\frac{6!}{2\times 2\times 2}=\frac{6\times 5\times 4\times 3\times 2}{2\times 2\times 2}\] \[=6\times 5\times 3\]

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

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

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

  65. mridrik
    • 2 years ago
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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

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

  67. satellite73
    • 2 years ago
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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 \[\frac{1}{7}\]

  68. mridrik
    • 2 years ago
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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

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

  70. satellite73
    • 2 years ago
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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

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

  72. mridrik
    • 2 years ago
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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

  73. satellite73
    • 2 years ago
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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

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

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

  76. satellite73
    • 2 years ago
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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

  77. mridrik
    • 2 years ago
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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

  78. mridrik
    • 2 years ago
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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

  79. satellite73
    • 2 years ago
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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

  80. satellite73
    • 2 years ago
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    yw

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

  82. satellite73
    • 2 years ago
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    lol

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