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

(See also)
(See also)
Line 7: Line 7:
 
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}
 +
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
 +
\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}}

Revision as of 18:00, 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

\begin{eqnarray} 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 \end{eqnarray}

See also

1991 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
Invalid username
Login to AoPS