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 4/9 the counter is put back in the bag. 4 more red counters and 2 more green counters are put in the bag. a counter is taken from the bag. the probability that the counter is green is 10/23 find the number of red counters and the number of green counters that were in the bag originally. your final line must say, ... red and ... green counters.

Answer 1

The number of red counters and the number of green counters that were in the bag originally is: 10 red and 8 green counters.

How to calculate the probability of an event?

Suppose that there are finite elementary events in the sample space of the considered experiment, and all are equally likely.

Then, suppose we want to find the probability of an event E.

Then, its probability is given as

$P(E) = \dfrac{\text{Number of favorable cases}}{\text{Number of total cases}} = \dfrac{n(E)}{n(S)}$

where favorable cases are those elementary events who belong to E, and total cases are the size of the sample space.

For this case, let we had:

R numbers of red counters, and G number of green counters originally.

Then, total counters in the bag originally = R +G

One counter out of R+G counters can be chosen in R+G ways. But one green counter out of R+G counters can be chosen in G ways.

Thus, probability of taking green counter out in one random picking is:

$\dfrac{G}{R + G}$

It is given that this probability is 4/9. Therefore, we get:

$\dfrac{G}{R+G} = \dfrac{4}{9}\\\\\\9G = 4R + 4G\\\\5G = 4R$(first equation)

When 4 more red counters and 2 more green counters were added, then:

• Number of red counters in modified bag = R + 4
• Number of green counters in modified bag = G + 2
• Number of total counters in modified bag = R + 4 + G+ 2 = R + G + 6

One counter out of R+G+6 counters can be chosen in R+G+6 ways. But one green counter out of R+G+6 counters can be chosen in G+2 ways (since there are G+2 green counters in the considered bag) ways.

Thus, probability of taking green counter out in one random picking is:

$\dfrac{G+2}{R + G+6}$

It is given that this probability is 10/23. Therefore, we get:

$\dfrac{G+2}{R+G+6} = \dfrac{10}{23}\\\\\\23G + 46 = 10R + 10G + 60\\\\13G = 10R + 14$(second equation)

Thus, we got a system of two linear equations as:

$5G = 4R\\13G = 10R + 14$

From first equation, getting the value of G in terms of R,

$G = \dfrac{4R}{5}$

Now, substituting this with G in the second equation, we get:

$\\13\left(\dfrac{4R}{5} \right) =10R + 20\\\\52R = 50R + 20\\2R = 20\\R = 10$

From this value of R, and the value of G in terms of R, we get:

$G = \dfrac{4R}{5}\\\\ = \dfrac{4 \times 10}{5} = 8$

Thus, R = 10 and G = 8.

Thus, the number of red counters and the number of green counters that were in the bag originally is: 10 red and 8 green counters.

Learn more about probability here:

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