John has a empty box. He puts some red counters and some blue counters. The ratio of the number of red counters to blue counters is 1:4 . Linda takes at r

Thanks Chris,

It was a slightly unusual question, I liked it too :)

Hi guest :)

John has a empty box. He puts some red counters and some blue counters. The ratio of the number of red counters to blue counters is 1:4. Linda takes at random 2 counters. The probability that she took 2 red counters is 6/155. How many red counters did John put in the box?

Let there be x red counters, then there will be 4x blue counters which equals 5x counters altogether in the box.

so the probability that the first counter will be red is     \(\frac{x}{5x}=\frac{1}{5}\)

Now there are x-1 red counters left out of   5x-1 counters in total so 

the probablility that the second counter is red is    \(\frac{x-1}{5x-1}\)

\(P( both\; red) = \frac{1}{5}\times \frac{(x-1)}{(5x-1)}=\frac{6}{155}\\~\\ \begin{align} \frac{(x-1)}{5(5x-1)}&=\frac{6}{155}\\ (x-1)*155&=6*5(5x-1)\\ 155x-155&=150x-30\\ 5x&=125\\ x&=25\\ \end{align}\)

So there was 25 red marbles in the box :)

Very nice, Melody...I like this one...!!!

So I expected.