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 11 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?
 11 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
 11 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
 11 months ago
Best ResponseYou've already chosen the best response.0i think its correct not solved but just what u wrote

nickersia
 11 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
 11 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
 11 months ago
Best ResponseYou've already chosen the best response.1your last step is wrong..

TwoPointInfinity
 11 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
 11 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
 11 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
 11 months ago
Best ResponseYou've already chosen the best response.1Hence probability of both happening is the product, not sum!

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

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

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

nickersia
 11 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
 11 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
 11 months ago
Best ResponseYou've already chosen the best response.0your quadratic equation is correct...your final solution is not

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