Answers:
12 red and 9 green counters originally
=======================================================
Explanation:
r = number of red counters originally in the bag
g = number of green counters originally in the bag
We have g number of green counters out of g+r counters total. The probability of picking green is g/(g+r). We set it equal to 3/7 as this is the stated probability of picking green.
Let's solve for g
g/(g+r) = 3/7
7g = (g+r)*3 ... cross multiply
7g = 3g+3r
7g-3g = 3r
4g = 3r
g = 3r/4
g = (3/4)r
g = 0.75r
We'll use this equation later.
------------------------------------------
Whatever counter is selected is placed back into the bag, or it is replaced.
Now we add 2 more red and 3 more green counters.
The green goes from g to g+3, while the total goes from g+r to g+r+2+3 = g+r+5
The probability of picking green after these counters are added in is now (g+3)/(g+r+5) which is the same as 6/13.
We get the equation
(g+3)/(g+r+5) = 6/13
We'll plug in the previous equation we found (when we isolated g), and solve for r like so
(g+3)/(g+r+5) = 6/13
(0.75r+3)/(0.75r+r+5) = 6/13
(0.75r+3)/(1.75r+5) = 6/13
13(0.75r+3) = 6(1.75r+5) ... cross multiply
9.75r+39 = 10.5r+30
9.75r-10.5r = 30-39
-0.75r = -9
r = -9/(-0.75)
r = 12
There were originally 12 red counters in the bag
g = 0.75*r
g = 0.75*12
g = 9
And there were originally 9 green counters in the bag.
This would mean there were 9+12 = 21 counters total. The probability of picking green is 9/21 = (3*3)/(3*7) = 3/7 which matches with what the instructions are saying.
If we added 2 red and 3 green, then we have 9+3 = 12 green and 21+2+3 = 26 total. The probability of picking green now is 12/26 = (2*6)/(2*13) = 6/13. This also matches with what the instructions mention. Therefore, the answers have been confirmed.
