The question is incomplete! Complete question along with answer and step by step explanation is provided below.
Question:
There are only r red counters and g green counters in a bag. A counter is taken at random from the bag. The probability that the counter is green is 3/7 The counter is put back in the bag. 2 more red counters and 3 more green counters are put in the bag. A counter is taken at random from the bag. The probability that the counter is green is 6/13
Find the number of red counters and the number of green counters that were in the bag originally.
Answer:
Total number of green counters = 9
Total number of red counters = 12
Step-by-step explanation:
Recall that probability is given by
P = number of desired events/total number of events
The probability that the counter is green is 3/7
P(green) = 3/7 = 3x/7x
Where 3x is the number of green counters
7x is the total number of counters
So then red counters are
red counters = 7x - 3x = 4x
4x is the number of red counters
P(green) = 3/7 = 3x/3x + 4x
The counter is put back in the bag. 2 more red counters and 3 more green counters are put in the bag. The probability that the counter is green is 6/13
So after addition of 3 green and 2 red new counters,
P(green) = 6/13 = (3x + 3)/(3x + 4x + 3 + 2)
Now solve for x
6/13 = (3x + 3)/(3x + 4x + 3 + 2)
6/13 = (3x + 3)/(7x + 5)
6(7x + 5) = 13(3x + 3)
42x + 30 = 39x + 39
42x - 39x = 39 - 30
3x = 9
x = 9/3
x = 3
So total number of green counters are
green counters = 3x = 3*3 = 9
So total number of red counters are
red counters = 4x = 4*3 = 12
