Fdtxjr

Two marbles are randomly selected without replacement from a bag
containing blue and green marbles. Probability of drawing both blue is
$frac16$. If three marbles are drawn, then the probability all
three are blue is $frac121$. What is the fewest number of marbles
that must have been in the bag before any were drawn?

I use the criteria to make two equations:
$$fracxycdotfracx-1y-1=frac16$$
for drawing 2 marbles$$fracxycdotfracx-1y-1cdotfracx-2y-2=frac121$$
Where $x$ is the number of blue marbles and $y$ is the total number of marbles.

Simplifying these equations gives:
$$6x^2-6x=y^2-y$$ for the first equation and $$21x^3-63x^2+42x=y^3-3y^2+2y$$

I can't seem to solve this set of equations.

Is my approach correct? If so, how should I continue? If not, how would one solve this problem?

Thanks! Your help is appreciated!

Max0815

Let's look at the second equation:

$$underbracedfracxycdot dfracx-1y-1_dfrac16cdot dfracx-2y-2 = dfrac121$$

The first two terms are the same as the LHS of your first equation. So, we can substitute.

$$dfracx-2y-2 = dfrac621$$

Reducing the RHS to lowest terms gives:

$$dfracx-2y-2 = dfrac27$$

There you have $x-2=2, y-2=7$

So, $x=4, y=9$ should be the smallest possible values for which the problem should hold true (if there were smaller values, it would contradict $dfrac27$ being in lowest terms).

Divide the second equation by the first:

$$fracx-2y-2 = frac621 = frac27$$

Which when simplified gives:

$$7x = 2y + 10$$

What are the smallest integer values of x and y that solve this equation? Well, observe that a possible solution is $x = 2, y = 2$, but this is clearly wrong since $x = y$ in this case. Since $x$ cannot be odd, we go on to the next number, which is 4. Luckily, we can observe that a solution will be given by: $$x = 4, y = 9$$

which happens to be the correct answer.

As Sean Lee and InterstellarProbe note, dividing the second equation by the first leaves

$$x-2over y-2=6over21=2over7$$

It follows that $x-2=2k$ and $y-2=7k$ for some $knot=0$. Plugging $x=2k+2$ and $y=7k+2$ into $6x^2-6x=y^2-y$, we have

$$6(2k+2)^2-6(2k+2)=(7k+2)^2-(7k+2)$$

which expands and simplifies initially to

$$24k^2+36k+12=49k^2+21k+2$$

and then to

$$25k^2-15k-10=5(5k+2)(k-1)=0$$

This has two solutions, $k=1$ and $k=-2/5$. The first gives a meaningful solution in integers, $x=4$, $y=9$. The second gives a meaningless non-solution, $x=1/5$, $y=-4/5$. Thus $9$ marbles (with $4$ of them blue) is not only the fewest number of marbles you can have to start with, it's the only number.