2002 AMC 12P Problems/Problem 15

The following problem is from both the 2002 AMC 12P #15 and 2002 AMC 10P #17, so both problems redirect to this page.

Problem

There are $1001$ red marbles and $1001$ black marbles in a box. Let $P_s$ be the probability that two marbles drawn at random from the box are the same color, and let $P_d$ be the probability that they are different colors. Find $|P_s-P_d|.$

$\text{(A) }0 \qquad \text{(B) }\frac{1}{2002} \qquad \text{(C) }\frac{1}{2001} \qquad \text{(D) }\frac {2}{2001} \qquad \text{(E) }\frac{1}{1000}$

Solution

First we find the value of $P_s$. Note that whatever color we choose on our first marble, there are exactly $1000$ of $2001$ marbles remaining that match that color. Therefore, $P_s = \frac {1000}{2001}$.

Now we find the value of $P_d$. Again, the actual color of the first marble does not matter, since there are always exactly $1001$ of $2001$ marbles remaining that don't match that color. Therefore, $P_d = \frac{1001}{2001}$.

The value of $|P_s - P_d|$ is therefore $\frac {|1000-1001|}{2001} = \boxed {\text{(C) }\frac{1}{2001}}$.


~Minor edits from Astro2010~

See also

2002 AMC 10P (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2002 AMC 12P (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png