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> | ||
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== Solution == | == Solution == | ||
| − | + | First we find the value of <math>P_s</math>. Note that whatever color we choose on our first marble, there are exactly <math>1000</math> of <math>2001</math> marbles remaining that match that color. Therefore, <math>P_s = \frac {1000}{2001}</math>. | |
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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 don't match that color. Therefore, <math>P_d = \frac{1001}{2001}</math>. | ||
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| + | The value of <math>|P_s - P_d|</math> is therefore <math>\frac {|1000-1001|}{2001} = \boxed {\text{(C) }\frac{1}{2001}}</math>. | ||
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| + | ~Minor edits from [[User:Astro2010|Astro2010]]~ | ||
== See also == | == See also == | ||
| + | {{AMC10 box|year=2002|ab=P|num-b=16|num-a=18}} | ||
{{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}} | ||
Latest revision as of 15:59, 14 August 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 
red marbles and black marbles in a box. Let 
be the probability that two marbles drawn at random from the box are the same color, and let 
be the probability that they are different colors. Find 
Solution
First we find the value of . Note that whatever color we choose on our first marble, there are exactly 
of 
marbles remaining that match that color. Therefore, 
.
Now we find the value of . Again, the actual color of the first marble does not matter, since there are always exactly of marbles remaining that don't match that color. Therefore, 
.
The value of 
is therefore 
.
~Minor edits from Astro2010~
See also
| 2002 AMC 10P (Problems • Answer Key • Resources) | ||
| 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 (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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
