anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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karatechopper
  • karatechopper
ok well first tell me how many marbles do u have all together?
anonymous
  • anonymous
10
amistre64
  • amistre64
yeah, satellite would definnantly be able to breeze by on this one

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.Sam.
  • .Sam.
@satellite73
anonymous
  • anonymous
Yes he is very good with probability I've noticed
amistre64
  • amistre64
rrbbgg 5/10 + 5/10 + 2/10 + 2/10 + 3/10 + 3/10 would be my guess but im sure its wrong
anonymous
  • anonymous
ok this is with replacement right?
amistre64
  • amistre64
i was oring, instead of anding ... among other fauxpauxes
anonymous
  • anonymous
bag has ten marbles and the number of ways to draw six out of ten is \[\dbinom{10}{6}=\frac{10\times 9\times 8\times 7}{4\times 3\times 2}=210\]
anonymous
  • anonymous
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 \[\dbinom{2}{2}=1\]
anonymous
  • anonymous
so your answer is \[\frac{\dbinom{5}{2}\times\dbinom{3}{2}\times \dbinom{2}{2}}{\dbinom{10}{6}}\] \[\frac{10\times 3\times 1}{210}=\frac{1}{7}\]
anonymous
  • anonymous
That's incorrect
anonymous
  • anonymous
way too big according to mathcounts
anonymous
  • anonymous
yes it certainly is isnt it!
anonymous
  • anonymous
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?
anonymous
  • anonymous
site clogged up one me
anonymous
  • anonymous
Do you see my last post on your computer?
anonymous
  • anonymous
yeah i messed up and did the problem as if it was without replacement
anonymous
  • anonymous
Can you show me how to do it using "my way" if there were no replacements?
anonymous
  • anonymous
ok sorry it took a while i had to think
anonymous
  • anonymous
here we go. rr gg bb has probability \[\frac{5}{10}\times \frac{5}{10}\times\frac{3}{10}\times \frac{3}{10}\times \frac{2}{10}\times \frac{2}{10}=\frac{3^3}{1000}\]
anonymous
  • anonymous
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
anonymous
  • anonymous
so our final answer is \[\frac{3^2\times 5\times 2\times 3^2}{10000}\] \[=\frac{81}{1000}\]
anonymous
  • anonymous
if there are no replacements the answer i wrote above would work (way above) to do it "your way" we would compute as follows
anonymous
  • anonymous
\[\frac{5\times 4\times 3\times 2\times 2\times 1}{10\times 9\times 8\times 7\times 6\times 5}\times 3\times 5\times 2\times 3\] 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
anonymous
  • anonymous
sorry i messed up the first time, i didn't read it correctly
anonymous
  • anonymous
It's all right, but can you explain your last post a little bit better please
anonymous
  • anonymous
i can try
anonymous
  • anonymous
first method i think is easier, the one where i wrote \[\frac{\dbinom{5}{2}\times\dbinom{3}{2}\times \dbinom{2}{2}}{\dbinom{10}{6}}=\frac{10\times 3\times 1}{210}=\frac{1}{7}\]
anonymous
  • anonymous
I don't really understand that one either :(
anonymous
  • anonymous
ok that one is easy to explain
anonymous
  • anonymous
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
anonymous
  • anonymous
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 \[\frac{30}{210}=\frac{1}{7}\]
anonymous
  • anonymous
Okay I see where you get the numbers but I don't see how that gives you the probability
anonymous
  • anonymous
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
anonymous
  • anonymous
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
anonymous
  • anonymous
now back to your method. we compute the probability we get exactly the sequence rr bb gg
anonymous
  • anonymous
right I understand that but I dont understand how that gives you the probability that 2 of each color is drawn w/o replacement
anonymous
  • anonymous
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.
anonymous
  • anonymous
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 \[\frac{1}{5}\]
anonymous
  • anonymous
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
anonymous
  • anonymous
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
anonymous
  • anonymous
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
anonymous
  • anonymous
wait, hold on stop for a moment, un momento!,
anonymous
  • anonymous
ok
anonymous
  • anonymous
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.
anonymous
  • anonymous
ok first is red there are 5 red, 10 total, so probability first is red is 1/2 yes?
anonymous
  • anonymous
yes.
anonymous
  • anonymous
then we compute the probability that second is red GIVEN that first is red. that is 4/9 right?
anonymous
  • anonymous
so probability that first is red AND second is red is \[\frac{5}{10}\times \frac{4}{9}\]
anonymous
  • anonymous
1/2 * 4/9, yes
anonymous
  • anonymous
ok now third is blue given first two are red. 3 blues, 8 left, so \[\frac{3}{8}\]
anonymous
  • anonymous
likewise fourth this blue given first red, second red, third blue is \[\frac{2}{7}\]
anonymous
  • anonymous
yes
anonymous
  • anonymous
fifth green given first red, second red, third blue, fourth blue \[\frac{2}{6}\]
anonymous
  • anonymous
then 1/5 i gotcha
anonymous
  • anonymous
and finally sixth is green given blah blah is \[\frac{1}{5}\] right
anonymous
  • anonymous
now we compute the probabilty that all this occurs by multiplyiing it all together
anonymous
  • anonymous
\[\frac{5}{10}\times \frac{4}{9}\times \frac{3}{8}\times\frac{2}{7}\times \frac{2}{6}\times \frac{1}{5}\] gives the probability we get R R B B G G in that order
anonymous
  • anonymous
but order does not matter, we just want 2 red, 2 blue and 2 green
anonymous
  • anonymous
so now the question is, how many different arrangments of r r b b g g are there? and the answer is \[\frac{6!}{2\times 2\times 2}=\frac{6\times 5\times 4\times 3\times 2}{2\times 2\times 2}\] \[=6\times 5\times 3\]
anonymous
  • anonymous
so multiply by 90 right. 6!/2!^3
anonymous
  • anonymous
i get 90, yes
anonymous
  • anonymous
each combination has same probability, (you can check this) so you multiply the number above by 90
anonymous
  • anonymous
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
anonymous
  • anonymous
yes they give the same thing it is true!
anonymous
  • anonymous
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 \[\frac{1}{7}\]
anonymous
  • anonymous
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
anonymous
  • anonymous
ok i don't want to annoy you with it but the counting principle is easy
anonymous
  • anonymous
if there are 10 ways to do one thing and 3 ways to do another, then there are 30 ways to do them together
anonymous
  • anonymous
so there are 30 ways to select 2 reds and 2 blues
anonymous
  • anonymous
I understand the counting principle fine, what I do not understand is how you used it to come up with your answer
anonymous
  • anonymous
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
anonymous
  • anonymous
so 30 ways to perform your task and 210 possible ways to choose 6 out of 10
anonymous
  • anonymous
since they are presumed to be equally likely probability is 30/210
anonymous
  • anonymous
counting prinicple tells you the number of ways you can do something. if they are equally likely then probabilty is simple enough
anonymous
  • anonymous
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
anonymous
  • anonymous
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
anonymous
  • anonymous
ok it works, although it is faily annoying to have to compute \[\frac{6!}{2^3}\] when it is not necessary
anonymous
  • anonymous
yw
anonymous
  • anonymous
i has calculator that does it for me, plus I know the formula
anonymous
  • anonymous
lol

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