# Math Help - two eays to solve

1. ## two eays to solve

What are two different ways that this problem can be solved. Using a permutation I know is one but what other ways are there?

The ratio of red marbles to green marbles in a container is 3 to 5. How many red marbles should you add to the container with n green marbles so that the probability of getting a red marble is 7/8? Give your answer in terms of n.

This is not a Permutations problem . . .

The ratio of red marbles to green marbles in a container is 3 to 5.
How many red marbles should you add to the container with n green marbles
so that the probability of getting a red marble is 7/8?
Presently, the ratio of red to green is: . $3:5$

If there are $n$ green marbles, the ratio is: . $\frac{3}{5}n:n$
. . and there are: . $\frac{3}{5}n+n\:=\:\frac{8}{5}n$ marbles in the container.

We will add $x$ red marbles.
Then there are: . $\frac{3}{5}n + x$ red marbles
. . and: . $\frac{8}{5}n + x$ marbles in the container.

Then we have: . $\frac{\dfrac{3}{5}n+x}{\dfrac{8}{5}n+x} \:=\:\frac{7}{8}\quad\Rightarrow\quad \frac{3n+5x}{8x+5x} \:=\:\frac{7}{8}$

. . $8(3n+5x) \:=\:7(8n+5x)\quad\Rightarrow\quad 24n + 40x \:=\:56n + 35x$

. . $5x \:=\:32n\quad\Rightarrow\quad\boxed{ x \:=\:\frac{32}{5}n}$

3. ## Another way?

Could there be a different way to solve this?