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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.
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.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0choices: 21/304 21/449 56/429 70/429 123/856

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0two 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
 5 years ago
Best ResponseYou've already chosen the best response.0hope 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: \[\frac{8}{13}\times \frac{3}{12}\times\frac{7}{11}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 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
 5 years ago
Best ResponseYou've already chosen the best response.0ok , i worked it out and got 168/17160. thats not an option either : ( helllpppp lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok 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
 5 years ago
Best ResponseYou've already chosen the best response.0oh damn now i see my mistake. it is not 3 green marbles it is 5!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry let me try again. \[3\times \frac{8}{13}\times \frac{7}{12}\times \frac{5}{11}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, let me know what u got : ) ur the best!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i get \[\frac{840}{1716}=\frac{70}{143}\] sorry for messing you up.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0same if you compute \[\frac{\dbinom{8}{2}\times \dbinom{5}{1}}{\dbinom{13}{3}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0does it matter which way you do it? or do you just have to get the answer?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats still not a choice : ( this os no bueno lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok 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
 5 years ago
Best ResponseYou've already chosen the best response.0we just have to have the answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i get \[\frac{70}{143}\] again doing it a different way. is it possible there is a mistake in the answers?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the 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
 5 years ago
Best ResponseYou've already chosen the best response.0that 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
 5 years ago
Best ResponseYou've already chosen the best response.0the 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
 5 years ago
Best ResponseYou've already chosen the best response.0ok ill just write that, thank you so much : ) sorry it stumped u too lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok ill just write that, thank you so much : ) sorry it stumped u too lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that 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
 5 years ago
Best ResponseYou've already chosen the best response.0i think there is a typo in the answers. i will check if any of the answers correspond to two green and one red.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ah 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.

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0I also got 70/143 but not so sure since I am weak on probability :).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whew i thought i was going nuts. i just needed another pair of "eyes" thnx

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Notice that \((70/143)/3=70/429\). So I am still thinking that my answer might be wrong.
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