Difference between revisions of "1991 AIME Problems/Problem 13"

(See also)
(Solution)
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The probability <math>P</math> that when two socks are drawn without replacement, both are red or both are blue is given by
 
The probability <math>P</math> that when two socks are drawn without replacement, both are red or both are blue is given by
  
\begin{eqnarray}
+
<math>
 
P=\frac{r(r-1)}{(r+b)(r+b-1)}+\frac{b(b-1)}{(r+b)(r+b-1)}=\frac{r(r-1)+b(b-1)}{t(t-1)}=\frac{1}{2}
 
P=\frac{r(r-1)}{(r+b)(r+b-1)}+\frac{b(b-1)}{(r+b)(r+b-1)}=\frac{r(r-1)+b(b-1)}{t(t-1)}=\frac{1}{2}
\nonumber
+
<math>
\end{eqnarray}
+
 
 
== See also ==
 
== See also ==
{{AIME box|year=1991|num-b=12|num-a=14}}
+
{{AIME box|year=1991|num-b=12|num-a=14}}</math>

Revision as of 18:02, 18 April 2007

Problem

A drawer contains a mixture of red socks and blue socks, at most 1991 in all. It so happens that, when two socks are selected randomly without replacement, there is a probability of exactly $\displaystyle \frac{1}{2}$ that both are red or both are blue. What is the largest possible number of red socks in the drawer that is consistent with this data?

Solution

Let $r$ and $b$ denote the number of red and blue socks, respectively. Also, let $t=r+b$.

The probability $P$ that when two socks are drawn without replacement, both are red or both are blue is given by

$P=\frac{r(r-1)}{(r+b)(r+b-1)}+\frac{b(b-1)}{(r+b)(r+b-1)}=\frac{r(r-1)+b(b-1)}{t(t-1)}=\frac{1}{2} <math>

== See also == {{AIME box|year=1991|num-b=12|num-a=14}}$ (Error compiling LaTeX. ! Missing $ inserted.)

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