A box contains R red balls, B blue balls, and no other balls. One ball is removed and set aside, and then a second ball is removed. On each draw, each ball in the box is equally likely to be removed. The probability that both of these balls are red is 2/7 . The probability that exactly one of these balls is red is 1/2 . Determine the pair (R, B).
+2862 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!
Well, If you have R red balls out of a total (R + B), then when you draw the first one(assuming it is R), then don't you have left (R - 1) / (R + B - 1) for the 2nd draw?. Use "concrete numbers" to illustrate the point: Suppose you have 10 Red ball and 15 Blue balls. The first draw will be: 10 / [10+15]=10/25. The 2nd draw will have: 10 - 1/25 -1 =9/24. Then the probability of 2 Red ball will be: 10/25 x 9/24.
+2862 oohhhhh, ok
guest's way of determining probability + my strategy of solving using equations = your answer
