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 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 Find the number of red counters and the number of green counters that were in the bag originally. ( 5 marks)

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