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Difference between revisions of "2002 AMC 12P Problems/Problem 15"

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{{duplicate|[[2002 AMC 12P Problems|2002 AMC 12P #15]] and [[2002 AMC 10P Problems|2002 AMC 10P #17]]}}
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== Problem ==
 
== Problem ==
 
There are <math>1001</math> red marbles and <math>1001</math> black marbles in a box. Let <math>P_s</math> be the probability that two marbles drawn at random from the box are the same color, and let <math>P_d</math> be the probability that they are different colors. Find <math>|P_s-P_d|.</math>
 
There are <math>1001</math> red marbles and <math>1001</math> black marbles in a box. Let <math>P_s</math> be the probability that two marbles drawn at random from the box are the same color, and let <math>P_d</math> be the probability that they are different colors. Find <math>|P_s-P_d|.</math>

Revision as of 16:41, 14 July 2024

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 match that color. Therefore, $P_d = \frac{1001}{2001}$.

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

See also

2002 AMC 12P (Problems • Answer Key • Resources)
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

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