a bag of marbles contains 8 red marbles and 5 green marbles. 3 marbles are drawn at random at the same time. find the probabilty that two are red and one is green.

- anonymous

- jamiebookeater

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- anonymous

choices:
21/304
21/449
56/429
70/429
123/856

- anonymous

two ways to do this. Compute the probability of getting (R R G) in that order, then multiply by 3 since you can also get (R G R) or (G R R) and they are all what you want.
probability that the first one is red and the second one is red and the third one is green is
\[\frac{8}{13}\times \frac{7}{12}\times \frac{3}{11}\]

- anonymous

hope it is clear why those numbers are correct. and hope it is also clear that if you compute the probability you get (R G R) you get the same number:
\[\frac{8}{13}\times \frac{3}{12}\times\frac{7}{11}\]

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## More answers

- anonymous

hmmm..

- anonymous

so answer should be \[3\times \frac{8}{13}\times \frac{7}{12}\times \frac{3}{11}\]
hmm indeed. i get \[\frac{504}{1716}=\frac{42}{143}\]
which you do not have. let me see if i made calculation error or a logic one.

- anonymous

ok , i worked it out and got 168/17160. thats not an option either : ( helllpppp lol

- anonymous

ok i did it another way and got the same answer, so i am going to stick to it. i computed
\[\frac{\dbinom{8}{2}\times \dbinom{3}{1}}{\dbinom{13}{3}}\] and got the same thing.

- anonymous

oh damn now i see my mistake. it is not 3 green marbles it is 5!

- anonymous

sorry let me try again.
\[3\times \frac{8}{13}\times \frac{7}{12}\times \frac{5}{11}\]

- anonymous

ok, let me know what u got : ) ur the best!

- anonymous

i get \[\frac{840}{1716}=\frac{70}{143}\]
sorry for messing you up.

- anonymous

same if you compute
\[\frac{\dbinom{8}{2}\times \dbinom{5}{1}}{\dbinom{13}{3}}\]

- anonymous

does it matter which way you do it? or do you just have to get the answer?

- anonymous

thats still not a choice : ( this os no bueno lol

- anonymous

ok i try one more time. make sure i have it right: 13 marbles total. 8 red. 5 green. you select 3. you want 2 red and 1 green. yes?

- anonymous

we just have to have the answer

- anonymous

yes thats right : )

- anonymous

i get \[\frac{70}{143}\] again doing it a different way. is it possible there is a mistake in the answers?

- anonymous

the number of ways you can choose 3 marbles from a set of 13 is
\[\dbinom{13}{3}=\frac{13\times 12\times 11}{3\times 2}=13\times 2\times 11=286\]

- anonymous

that is your denominator.
the number of ways you can choose 2 marbles from a set of 8 is
\[\dbinom{8}{2}=\frac{8\times 7}{2}=4\times 7 = 28\]

- anonymous

the number of ways to choose 1 marble from a set of 5 is just 5. so i get
\[\frac{28\times 5}{28}=\frac{140}{286}=\frac{70}{143}\]

- anonymous

ok ill just write that, thank you so much : ) sorry it stumped u too lol

- anonymous

ok ill just write that, thank you so much : ) sorry it stumped u too lol

- anonymous

that answer is right. i just computed the probability you get all three red, all three green, 2 green and 1 red, two red and one green. i can write them for you if you like. i verified that the numbers added to 1, so i am sure they are correct.

- anonymous

i think there is a typo in the answers. i will check if any of the answers correspond to two green and one red.

- anonymous

ah you found it. i made a mistake at the beginning when i used 3 instead of 5 for the number of green marbles. but then i fixed my mistake and even computed all the possible outcome and my second answer of \[\frac{70}{143}\] looks right to me.

- anonymous

ok : ) thanks again!

- watchmath

I also got 70/143 but not so sure since I am weak on probability :).

- anonymous

whew i thought i was going nuts. i just needed another pair of "eyes" thnx

- watchmath

Notice that \((70/143)/3=70/429\). So I am still thinking that my answer might be wrong.

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