Question

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?

Answer 1

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

Answer 2

10

Answer 3

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

Answer 4

@satellite73

Answer 5

Yes he is very good with probability I've noticed

Answer 6

rrbbgg

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

Answer 7

ok this is with replacement right?

Answer 8

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

Answer 9

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

Answer 10

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
\[\dbinom{2}{2}=1\]

Answer 11

so your answer is
\[\frac{\dbinom{5}{2}\times\dbinom{3}{2}\times \dbinom{2}{2}}{\dbinom{10}{6}}\]
\[\frac{10\times 3\times 1}{210}=\frac{1}{7}\]

Answer 12

That's incorrect

Answer 13

way too big according to mathcounts

Answer 14

yes it certainly is isnt it!

Answer 15

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?

Answer 16

site clogged up one me

Answer 17

Do you see my last post on your computer?

Answer 18

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

Answer 19

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

Answer 20

ok sorry it took a while i had to think

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

if there are no replacements the answer i wrote above would work (way above)
to do it "your way" we would compute as follows

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

i can try

Answer 29

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

Answer 30

I don't really understand that one either :(

Answer 31

ok that one is easy to explain

Answer 32

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

Answer 33

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
\[\frac{30}{210}=\frac{1}{7}\]

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

right I understand that but I dont understand how that gives you the probability that 2 of each color is drawn w/o replacement

Answer 39

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.

Answer 40

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
\[\frac{1}{5}\]

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

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

Answer 45

ok

Answer 46

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.

Answer 47

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

Answer 48

yes.

Answer 49

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

Answer 50

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

Answer 51

1/2 * 4/9, yes

Answer 52

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

Answer 53

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

Answer 54

yes

Answer 55

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

Answer 56

then 1/5 i gotcha

Answer 57

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

Answer 58

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

Answer 59

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

Answer 60

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

Answer 61

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

Answer 62

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

Answer 63

i get 90, yes

Answer 64

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

Answer 65

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

Answer 66

yes they give the same thing it is true!

Answer 67

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
\[\frac{1}{7}\]

Answer 68

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

Answer 69

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

Answer 70

if there are 10 ways to do one thing and 3 ways to do another, then there are 30 ways to do them together

Answer 71

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

Answer 72

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

Answer 73

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

Answer 74

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

Answer 75

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

Answer 76

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

Answer 77

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

Answer 78

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

Answer 79

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

Answer 80

yw

Answer 81

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

Answer 82

lol