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?

- anonymous

- jamiebookeater

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- karatechopper

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

- anonymous

10

- amistre64

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

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## More answers

- .Sam.

- anonymous

Yes he is very good with probability I've noticed

- amistre64

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

- anonymous

ok this is with replacement right?

- amistre64

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

- 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

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

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

That's incorrect

- anonymous

way too big according to mathcounts

- anonymous

yes it certainly is isnt it!

- 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

site clogged up one me

- anonymous

Do you see my last post on your computer?

- anonymous

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

- anonymous

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

- anonymous

ok sorry it took a while i had to think

- 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

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

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

- 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

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

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

- anonymous

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

- anonymous

i can try

- 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

I don't really understand that one either :(

- anonymous

ok that one is easy to explain

- 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

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

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

- 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

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

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

- 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

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

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

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

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

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

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

- anonymous

ok

- 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

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

- anonymous

yes.

- anonymous

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

- anonymous

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

- anonymous

1/2 * 4/9, yes

- anonymous

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

- anonymous

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

- anonymous

yes

- anonymous

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

- anonymous

then 1/5 i gotcha

- anonymous

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

- anonymous

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

- 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

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

- 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

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

- anonymous

i get 90, yes

- anonymous

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

- 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

yes they give the same thing it is true!

- 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

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

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

- 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

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

- anonymous

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

- 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

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

- anonymous

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

- anonymous

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

- 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

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

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

- anonymous

yw

- anonymous

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

- anonymous

lol

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