Question

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?

Answer 1

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?

Answer 2

i think its correct not solved but just what u wrote

Answer 3

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

Answer 4

it might be 12+(x-1)

Answer 5

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

Answer 6

your last step is wrong..

Answer 7

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

Answer 8

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

Answer 9

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)

Answer 10

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

Answer 11

\[P(A and B) = P(A).P(B provided A has happened)\]

Answer 12

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

Answer 13

yeh Mashy is correct , multiply the terms :)

Answer 14

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

Answer 15

try solving again

Answer 16

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

Answer 17

Sorry

Answer 18

your quadratic equation is correct...your final solution is not

Answer 19

Wrong typing in calculator, I got 6, thanks guys! :)