anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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?
rvc
  • rvc
i think its correct not solved but just what u wrote
anonymous
  • anonymous
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

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More answers

rvc
  • rvc
it might be 12+(x-1)
anonymous
  • anonymous
It makes no difference because at some point you'll have to get rid of the brackets :L
anonymous
  • anonymous
your last step is wrong..
TwoPointInfinity
  • TwoPointInfinity
I don't see this being possible. How can both marbles have the same probability if the first marble wasn't replaced?
anonymous
  • anonymous
I think u should multiply (x/12+x) *(x-1/11+x)=5/51 like this
anonymous
  • anonymous
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)
anonymous
  • anonymous
Hence probability of both happening is the product, not sum!
anonymous
  • anonymous
\[P(A and B) = P(A).P(B provided A has happened)\]
anonymous
  • anonymous
Oh, I got ye, I'll try it that way
anonymous
  • anonymous
yeh Mashy is correct , multiply the terms :)
anonymous
  • anonymous
\[\frac{ x ^{2}-x}{ x ^{2}+23x+132 }=\frac{ 5 }{ 51 }\] ... ... ... Leads me to x1=6.18 and x2=-2.57
Zarkon
  • Zarkon
try solving again
anonymous
  • anonymous
\[51x^{2}-51x=5x ^{2}+115x+660\] \[46x ^{2}-166x-660=0\] \[x _{1}=6.18 >----< x _{2}==2.57\]
anonymous
  • anonymous
Sorry
Zarkon
  • Zarkon
your quadratic equation is correct...your final solution is not
anonymous
  • anonymous
Wrong typing in calculator, I got 6, thanks guys! :)

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