Green Marbles

You walk up to a black jar containing fewer than 20 marbles. If you reach into the jar and randomly remove two marbles out of the jar without replacing the marbles, you have a 50 percent chance of getting two green marbles. How many green marbles are in the jar before you select the two marbles?
Source: mathcontest.olemiss.edu 11/28/2011

SOLUTION
Suppose there are three green balls labeled G_1,G_2,G_3 and two non-green balls labeled X_1,X_2 in the jar. We randomly remove two balls out of the jar and record the possible outcomes below:
G_1,G_2
G_1,G_3
G_1,X_1
G_1,X_2
G_2,G_3
G_2,X_1
G_2,X_2
G_3,X_1
G_3,X_2
X_1,X_2

Out of those ten possible outcomes we get three desirable outcomes, namely
G_1,G_2
G_1,G_3
G_2,G_3

Hence, the probability of getting two green marbles equals \frac{3}{10}.

In the above example, let n=5 be the total number of marbles (green and non-green) in the jar and let r=3 be the number of green marbles in the jar. Note that we do not care about the order of the two green marbles. We just want two green marbles.

Number of possible outcomes equals _nC_2=\,_5C_2=10

Number of desirable outcomes equals _rC_2=\,_3C_2=3

Probability of getting two green marbles equals

\frac{_rC_2}{_nC_2}=\frac{3}{10}=30\%

General Case
Let’s calculate the probability in the general case where n<20.

\frac{_rC_2}{_nC_2}=\frac{\frac{r!}{2!\left (r-2\right )!}}{\frac{n!}{2!\left (n-2\right )!}}

=\frac{r!}{2!\left (r-2\right )!}\times \frac{2!\left (n-2\right )!}{n!}

=\frac{r!}{\left (r-2\right )!}\times \frac{\left (n-2\right )!}{n!}

=\frac{r\left (r-1\right )\left (r-2\right )!}{\left (r-2\right )!}\times \frac{\left (n-2\right )!}{n\left (n-1\right )\left (n-2\right )!}

=\frac{r\left (r-1\right )}{n\left (n-1\right )}

We want the probability to be 50%:

\frac{r\left (r-1\right )}{n\left (n-1\right )}=\frac{1}{2}

Since all we know is that n<20, we will form products of pairs of consecutive positive integers from 1 through 19 and look for products that satisfy the ratio of 1 to 2. For example,
2\times 1=2
3\times 2=6
\cdots
We tabulate the results in the following table

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
  2 6 12 20 30 42 56 72 90 110 132 156 182 210 240 272 306 342

The table shows that the only products that satisfy the ratio are 6 and 12. We write
\frac{r\left (r-1\right )}{n\left (n-1\right )}=\frac{3\left (3-1\right )}{4\left (4-1\right )}=\frac{3\left (2\right )}{4\left (3\right )}=\frac{6}{12}=\frac{1}{2}

Thus, r=3 and there are 3 green balls in the jar.

Answer: 3.

Advertisements

About mvtrinh

Retired high school math teacher.
This entry was posted in Problem solving and tagged , , , . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s