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nickersia

  • 9 months ago

A bag contains 8 blue marbles, 4 red marbles and x white marbles. A marble is drawn at random and not replaced. A second marble is drawn at random. If the probability that both are white is 5/51, how many white marbles are in the bag?

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  1. nickersia
    • 9 months ago
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    What I did is: In bag at the start \[8+4+x=12+x\] Probability that first one is white: \[\frac{ x }{ 12+x}\] Now in the bag: \[8+4+x-1=11+x\] Probability that second one is white: \[\frac{ x-1 }{ 11+x }\] So \[\frac{ x }{ 12+x } + \frac{ x-1 }{ 11+x } =\frac{ 5 }{ 51 }\] Is that true?

  2. rvc
    • 9 months ago
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    i think its correct not solved but just what u wrote

  3. nickersia
    • 9 months ago
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    I just solved it for x, and I got x1=1.138 and x2=-11.51 I don't think that's correct, cause I guess I should be getting whole number... There can't be 1.138 marbles in the bag :S

  4. rvc
    • 9 months ago
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    it might be 12+(x-1)

  5. nickersia
    • 9 months ago
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    It makes no difference because at some point you'll have to get rid of the brackets :L

  6. Mashy
    • 9 months ago
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    your last step is wrong..

  7. TwoPointInfinity
    • 9 months ago
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    I don't see this being possible. How can both marbles have the same probability if the first marble wasn't replaced?

  8. moli1993
    • 9 months ago
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    I think u should multiply (x/12+x) *(x-1/11+x)=5/51 like this

  9. Mashy
    • 9 months ago
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    You first calculated the probability of picking a white ball then u calculated the probability of picking another white ball PROVIDED you have already picked one white ball (this is a conditional probability)

  10. Mashy
    • 9 months ago
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    Hence probability of both happening is the product, not sum!

  11. Mashy
    • 9 months ago
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    \[P(A and B) = P(A).P(B provided A has happened)\]

  12. nickersia
    • 9 months ago
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    Oh, I got ye, I'll try it that way

  13. moli1993
    • 9 months ago
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    yeh Mashy is correct , multiply the terms :)

  14. nickersia
    • 9 months ago
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    \[\frac{ x ^{2}-x}{ x ^{2}+23x+132 }=\frac{ 5 }{ 51 }\] ... ... ... Leads me to x1=6.18 and x2=-2.57

  15. Zarkon
    • 9 months ago
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    try solving again

  16. nickersia
    • 9 months ago
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    \[51x^{2}-51x=5x ^{2}+115x+660\] \[46x ^{2}-166x-660=0\] \[x _{1}=6.18 >----< x _{2}==2.57\]

  17. nickersia
    • 9 months ago
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    Sorry

  18. Zarkon
    • 9 months ago
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    your quadratic equation is correct...your final solution is not

  19. nickersia
    • 9 months ago
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    Wrong typing in calculator, I got 6, thanks guys! :)

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