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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.

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  1. anonymous
    • 5 years ago
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    choices: 21/304 21/449 56/429 70/429 123/856

  2. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    hmmm..

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

  7. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    oh damn now i see my mistake. it is not 3 green marbles it is 5!

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

  10. anonymous
    • 5 years ago
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    ok, let me know what u got : ) ur the best!

  11. anonymous
    • 5 years ago
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    i get \[\frac{840}{1716}=\frac{70}{143}\] sorry for messing you up.

  12. anonymous
    • 5 years ago
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    same if you compute \[\frac{\dbinom{8}{2}\times \dbinom{5}{1}}{\dbinom{13}{3}}\]

  13. anonymous
    • 5 years ago
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    does it matter which way you do it? or do you just have to get the answer?

  14. anonymous
    • 5 years ago
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    thats still not a choice : ( this os no bueno lol

  15. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    we just have to have the answer

  17. anonymous
    • 5 years ago
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    yes thats right : )

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

  19. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    ok ill just write that, thank you so much : ) sorry it stumped u too lol

  23. anonymous
    • 5 years ago
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    ok ill just write that, thank you so much : ) sorry it stumped u too lol

  24. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    ok : ) thanks again!

  28. watchmath
    • 5 years ago
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    I also got 70/143 but not so sure since I am weak on probability :).

  29. anonymous
    • 5 years ago
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    whew i thought i was going nuts. i just needed another pair of "eyes" thnx

  30. watchmath
    • 5 years ago
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    Notice that \((70/143)/3=70/429\). So I am still thinking that my answer might be wrong.

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