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

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Now we find the value of <math>P_d</math>. Again, the actual color of the first marble does not matter, since there are always exactly <math>1001</math> of <math>2001</math> marbles remaining that match that color. Therefore, <math>P_d = \frac{1001}{2001}</math>.
 
Now we find the value of <math>P_d</math>. Again, the actual color of the first marble does not matter, since there are always exactly <math>1001</math> of <math>2001</math> marbles remaining that match that color. Therefore, <math>P_d = \frac{1001}{2001}</math>.
  
−
The value of
+
The value of <math>|P_s - P_d|</math> is therefore <math>|\frac {1000-1001}{2001}|</math>.
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2002|ab=P|num-b=14|num-a=16}}
 
{{AMC12 box|year=2002|ab=P|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:15, 10 March 2024

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 $|\frac {1000-1001}{2001}|$.

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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