A Hard Probability Question????

Let ${\color{#D61F06}R}$ be the number of red marbles in the bag and let ${\color{#3D99F6}B}$ be the number of blue marbles in the bag. From the first drawing, we know that $\begin{aligned} \frac{{\color{#D61F06}R}}{{\color{#D61F06}R}+ {\color{#3D99F6}B}} &= \frac{1}{5} \\ {\color{#D61F06}R}+ {\color{#3D99F6}B} &= 5{\color{#D61F06}R} \end{aligned}$

From the second drawing, we know that $\begin{aligned} \frac{{\color{#D61F06}R} + 5}{{\color{#D61F06}R}+ {\color{#3D99F6}B} + 5} &= \frac{1}{3} \\ {\color{#D61F06}R}+ {\color{#3D99F6}B} + 5&= 3{\color{#D61F06}R} + 15 \\ 5{\color{#D61F06}R} + 5 &= 3{\color{#D61F06}R} + 15 \\ 2{\color{#D61F06}R} &= 10 \\ {\color{#D61F06}R} &= \boxed{5} \end{aligned}$

Simultaneous equations.

*Let r be the number of red marbles there originally were in the bag, and b be the number of blue marbles that were in the bag. *

we can say:

$\frac{r}{r+b}$ = $\frac{1}{5}$

Now after you add five red marbles, you're number of favourable outcomes (picking red) increases by five and so does your bag size.

So we can now say:

$\frac{r+5}{r+b+5}$ = $\frac{1}{3}$

simplifying equation two yields

$2r+10=b$

equating equation 1 and 2, we get

$2r+10=4r$

solving for r, which is the number of red marbles, we get

r=5