High School

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 \(\frac{4}{9}\). The counter is put back in the bag. Four more red counters and two more green counters are added to the bag. A counter is taken from the bag again. The probability that the counter is green is \(\frac{10}{23}\).

Find the number of red counters and the number of green counters that were originally in the bag.

Your final answer must state: "... red and ... green counters."

Answer :

The number of green and red counters are 35 red counters and 28 green counters

Probability is the likelihood for an event to occur. The certainty for an event to occur is 1 which is equivalent to 100% in percentage.

Probability is expressed as;

Probability = sample space /Total outcome

let x be the red counters And y be the green counters.

Total number of counter = x + y

The probability of taking green counters = y /(x + y)

4/9 = y /(x + y)

4x + 4y = 9y

4x = 5y

Four more red and two more greens are added to the counters

number of red counters = x + 4

number of green counters = y+ 2

Total counters = x + y + 6

Probability of taking green counter = y + 2/(x + y + 6)

10/23 = y + 2/(x + y + 6)

10x + 10y + 60 = 23y + 46

10x - 13y = -14

The two linear equation formed are

4x = 5y

10x - 13y = -14

x = (5/4)y

10(5/4)y - 13y = -14

(50/4)y - 13y = -14

50y - 52y = -56

- 2y = -56

y = -56/-2

y = 28 counters

x= 5/4 × 28

x = 5 × 7

x = 35 counters

Therefore, there are 35 red counters and 28 green counters in the bag

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