Difference between revisions of "2002 AMC 12P Problems/Problem 15"
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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>. | |
| + | |||
| + | 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 | ||
== 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 18:13, 10 March 2024
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 match that color. Therefore, 
.
The value of
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
