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nickersia
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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?
 8 months ago
 8 months ago
nickersia Group Title
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?
 8 months ago
 8 months ago

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nickersia Group TitleBest ResponseYou've already chosen the best response.0
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: \[8+4+x1=11+x\] Probability that second one is white: \[\frac{ x1 }{ 11+x }\] So \[\frac{ x }{ 12+x } + \frac{ x1 }{ 11+x } =\frac{ 5 }{ 51 }\] Is that true?
 8 months ago

rvc Group TitleBest ResponseYou've already chosen the best response.0
i think its correct not solved but just what u wrote
 8 months ago

nickersia Group TitleBest ResponseYou've already chosen the best response.0
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
 8 months ago

rvc Group TitleBest ResponseYou've already chosen the best response.0
it might be 12+(x1)
 8 months ago

nickersia Group TitleBest ResponseYou've already chosen the best response.0
It makes no difference because at some point you'll have to get rid of the brackets :L
 8 months ago

Mashy Group TitleBest ResponseYou've already chosen the best response.1
your last step is wrong..
 8 months ago

TwoPointInfinity Group TitleBest ResponseYou've already chosen the best response.0
I don't see this being possible. How can both marbles have the same probability if the first marble wasn't replaced?
 8 months ago

moli1993 Group TitleBest ResponseYou've already chosen the best response.0
I think u should multiply (x/12+x) *(x1/11+x)=5/51 like this
 8 months ago

Mashy Group TitleBest ResponseYou've already chosen the best response.1
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)
 8 months ago

Mashy Group TitleBest ResponseYou've already chosen the best response.1
Hence probability of both happening is the product, not sum!
 8 months ago

Mashy Group TitleBest ResponseYou've already chosen the best response.1
\[P(A and B) = P(A).P(B provided A has happened)\]
 8 months ago

nickersia Group TitleBest ResponseYou've already chosen the best response.0
Oh, I got ye, I'll try it that way
 8 months ago

moli1993 Group TitleBest ResponseYou've already chosen the best response.0
yeh Mashy is correct , multiply the terms :)
 8 months ago

nickersia Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{ x ^{2}x}{ x ^{2}+23x+132 }=\frac{ 5 }{ 51 }\] ... ... ... Leads me to x1=6.18 and x2=2.57
 8 months ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.0
try solving again
 8 months ago

nickersia Group TitleBest ResponseYou've already chosen the best response.0
\[51x^{2}51x=5x ^{2}+115x+660\] \[46x ^{2}166x660=0\] \[x _{1}=6.18 >< x _{2}==2.57\]
 8 months ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.0
your quadratic equation is correct...your final solution is not
 8 months ago

nickersia Group TitleBest ResponseYou've already chosen the best response.0
Wrong typing in calculator, I got 6, thanks guys! :)
 8 months ago
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