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 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?
 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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nickersia
 9 months ago
Best ResponseYou've already chosen the best response.0What 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?

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

nickersia
 9 months ago
Best ResponseYou've already chosen the best response.0I 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

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

Mashy
 9 months ago
Best ResponseYou've already chosen the best response.1your last step is wrong..

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

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

Mashy
 9 months ago
Best ResponseYou've already chosen the best response.1You 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)

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

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

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

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

nickersia
 9 months ago
Best 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

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

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

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