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 put in the bag. A counter is taken from the bag. 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 in the bag originally.

Question

High School

Answer :

ps3164771 ps3164771 · Answer 1 · 2024-08-21T11:26:23-04:00

To solve this problem, set up a system of equations and solve simultaneously for the number of red and green counters originally in the bag.

To solve this problem, we can set up a system of equations.

Let r be the number of red counters and g be the number of green counters originally in the bag.

According to the given information, the probability of getting a green counter on the first draw is 4/9.

So, the equation is (g/(r+g)) = 4/9.

After putting the counters back in the bag and adding more counters, the probability of getting a green counter on the second draw is 10/23.

So, the equation is ((g+2)/(r+g+6)) = 10/23.

Now, we can solve these equations simultaneously to find the values of r and g.

The solution to the system of equations is r = 5 and g = 9.