Question

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.

Answer 1

choices:
21/304

21/449

56/429

70/429

123/856

Answer 2

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
\[\frac{8}{13}\times \frac{7}{12}\times \frac{3}{11}\]

Answer 3

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:
\[\frac{8}{13}\times \frac{3}{12}\times\frac{7}{11}\]

Answer 4

hmmm..

Answer 5

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.

Answer 6

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

Answer 7

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.

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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?

Answer 16

we just have to have the answer

Answer 17

yes thats right : )

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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.

Answer 25

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

Answer 26

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.

Answer 27

ok : ) thanks again!

Answer 28

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

Answer 29

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

Answer 30

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