Difference between revisions of "1991 AIME Problems/Problem 13"
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<math> | <math> | ||
− | P=\frac{r(r-1)}{(r+b)(r+b-1)}+\frac{b(b-1)}{(r+b)(r+b-1)}=\frac{r(r-1)+ | + | P=\frac{r(r-1)}{(r+b)(r+b-1)}+\frac{b(b-1)}{(r+b)(r+b-1)}=\frac{r(r-1)+(t-r)(t-r-1)}{t(t-1)}=\frac{1}{2}. |
</math> | </math> | ||
+ | Solving the quadratic equation for <math>r</math> in terms of <math>t</math>, one obtains | ||
== See also == | == See also == | ||
{{AIME box|year=1991|num-b=12|num-a=14}}</math> | {{AIME box|year=1991|num-b=12|num-a=14}}</math> |
Revision as of 19:05, 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 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 and
denote the number of red and blue socks, respectively. Also, let
.
The probability that when two socks are drawn without replacement, both are red or both are blue is given by
Solving the quadratic equation for
in terms of
, one obtains
See also
1991 AIME (Problems • Answer Key • Resources) | ||
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 |
</math>