## mridrik 3 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?

1. karatechopper

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

2. mridrik

10

3. amistre64

yeah, satellite would definnantly be able to breeze by on this one

4. .Sam.

@satellite73

5. mridrik

Yes he is very good with probability I've noticed

6. amistre64

rrbbgg 5/10 + 5/10 + 2/10 + 2/10 + 3/10 + 3/10 would be my guess but im sure its wrong

7. satellite73

ok this is with replacement right?

8. amistre64

i was oring, instead of anding ... among other fauxpauxes

9. satellite73

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

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

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

That's incorrect

13. mridrik

way too big according to mathcounts

14. satellite73

yes it certainly is isnt it!

15. mridrik

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

site clogged up one me

17. mridrik

Do you see my last post on your computer?

18. satellite73

yeah i messed up and did the problem as if it was without replacement

19. mridrik

Can you show me how to do it using "my way" if there were no replacements?

20. satellite73

ok sorry it took a while i had to think

21. satellite73

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

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

so our final answer is $\frac{3^2\times 5\times 2\times 3^2}{10000}$ $=\frac{81}{1000}$

24. satellite73

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

$\frac{5\times 4\times 3\times 2\times 2\times 1}{10\times 9\times 8\times 7\times 6\times 5}\times 3\times 5\times 2\times 3$ 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

sorry i messed up the first time, i didn't read it correctly

27. mridrik

It's all right, but can you explain your last post a little bit better please

28. satellite73

i can try

29. satellite73

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

I don't really understand that one either :(

31. satellite73

ok that one is easy to explain

32. satellite73

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

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

Okay I see where you get the numbers but I don't see how that gives you the probability

35. satellite73

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

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

now back to your method. we compute the probability we get exactly the sequence rr bb gg

38. mridrik

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

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

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

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

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

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

wait, hold on stop for a moment, un momento!,

45. satellite73

ok

46. mridrik

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

ok first is red there are 5 red, 10 total, so probability first is red is 1/2 yes?

48. mridrik

yes.

49. satellite73

then we compute the probability that second is red GIVEN that first is red. that is 4/9 right?

50. satellite73

so probability that first is red AND second is red is $\frac{5}{10}\times \frac{4}{9}$

51. mridrik

1/2 * 4/9, yes

52. satellite73

ok now third is blue given first two are red. 3 blues, 8 left, so $\frac{3}{8}$

53. satellite73

likewise fourth this blue given first red, second red, third blue is $\frac{2}{7}$

54. mridrik

yes

55. satellite73

fifth green given first red, second red, third blue, fourth blue $\frac{2}{6}$

56. mridrik

then 1/5 i gotcha

57. satellite73

and finally sixth is green given blah blah is $\frac{1}{5}$ right

58. satellite73

now we compute the probabilty that all this occurs by multiplyiing it all together

59. satellite73

$\frac{5}{10}\times \frac{4}{9}\times \frac{3}{8}\times\frac{2}{7}\times \frac{2}{6}\times \frac{1}{5}$ gives the probability we get R R B B G G in that order

60. satellite73

but order does not matter, we just want 2 red, 2 blue and 2 green

61. satellite73

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

so multiply by 90 right. 6!/2!^3

63. satellite73

i get 90, yes

64. satellite73

each combination has same probability, (you can check this) so you multiply the number above by 90

65. mridrik

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

yes they give the same thing it is true!

67. satellite73

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

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

ok i don't want to annoy you with it but the counting principle is easy

70. satellite73

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

so there are 30 ways to select 2 reds and 2 blues

72. mridrik

I understand the counting principle fine, what I do not understand is how you used it to come up with your answer

73. satellite73

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

so 30 ways to perform your task and 210 possible ways to choose 6 out of 10

75. satellite73

since they are presumed to be equally likely probability is 30/210

76. satellite73

counting prinicple tells you the number of ways you can do something. if they are equally likely then probabilty is simple enough

77. mridrik

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

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

ok it works, although it is faily annoying to have to compute $\frac{6!}{2^3}$ when it is not necessary

80. satellite73

yw

81. mridrik

i has calculator that does it for me, plus I know the formula

82. satellite73

lol