## anonymous 5 years ago 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.

1. anonymous

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

2. 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 $\frac{8}{13}\times \frac{7}{12}\times \frac{3}{11}$

3. 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: $\frac{8}{13}\times \frac{3}{12}\times\frac{7}{11}$

4. anonymous

hmmm..

5. 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.

6. anonymous

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

7. 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.

8. anonymous

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

9. anonymous

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

10. anonymous

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

11. anonymous

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

12. anonymous

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

13. anonymous

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

14. anonymous

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

15. 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?

16. anonymous

we just have to have the answer

17. anonymous

yes thats right : )

18. anonymous

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

19. anonymous

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

20. anonymous

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

21. anonymous

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

22. anonymous

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

23. anonymous

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

24. 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.

25. 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.

26. 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.

27. anonymous

ok : ) thanks again!

28. watchmath

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

29. anonymous

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

30. watchmath

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