I tried to solve it, but it was too hard. This was my attempt, can someone tell me what I did wrong?
The probability that one red ball is selected:
\(\frac{r}{r+b}\).
With \(r\) being the number of red balls over \((r+b)\) , the total number of balls.
The probability that one blue ball is selected is similar:
\(\frac{b}{r+b}\).
Ok, so the problem states: "The probability that both of these balls are red is 2/7"
Lets make an equation for this:
\(\frac{r}{r+b}*\frac{r}{r+b}=\frac{2}{7}\)
The problem also states: "The probability that exactly one of these balls is red is 1/2"
Lets make an equation for this:
\(\frac{r}{r+b}*\frac{b}{r+b}=\frac{1}{2}\)
So now we have a system of equations:
\(\frac{r}{r+b}*\frac{r}{r+b}=\frac{2}{7}\)
\(\frac{r}{r+b}*\frac{b}{r+b}=\frac{1}{2}\)
We solve:(Simplify fraction multiplications)
\(\frac{r^2}{(r+b)^2}=\frac{2}{7}\)
\(\frac{rb}{(r+b)^2}=\frac{1}{2}\)
Now cross multiply:
\(7r^2=2(r+b)^2\)
\(2rb=(r+b)^2\)
Simplify:
\(\frac{7r^2}{2}=(r+b)^2\)
\(2rb=(r+b)^2\)
After this I got stuck, I determined the ratio between r and b, which is \(7r=4b\).
HelP!
