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