## anonymous 2 years 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?

1. 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?

2. rvc

i think its correct not solved but just what u wrote

3. 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

4. rvc

it might be 12+(x-1)

5. anonymous

It makes no difference because at some point you'll have to get rid of the brackets :L

6. anonymous

7. TwoPointInfinity

I don't see this being possible. How can both marbles have the same probability if the first marble wasn't replaced?

8. anonymous

I think u should multiply (x/12+x) *(x-1/11+x)=5/51 like this

9. 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)

10. anonymous

Hence probability of both happening is the product, not sum!

11. anonymous

$P(A and B) = P(A).P(B provided A has happened)$

12. anonymous

Oh, I got ye, I'll try it that way

13. anonymous

yeh Mashy is correct , multiply the terms :)

14. anonymous

$\frac{ x ^{2}-x}{ x ^{2}+23x+132 }=\frac{ 5 }{ 51 }$ ... ... ... Leads me to x1=6.18 and x2=-2.57

15. Zarkon

try solving again

16. anonymous

$51x^{2}-51x=5x ^{2}+115x+660$ $46x ^{2}-166x-660=0$ $x _{1}=6.18 >----< x _{2}==2.57$

17. anonymous

Sorry

18. Zarkon