Difference between revisions of "1986 AJHSME Problems/Problem 24"
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==Solution== | ==Solution== | ||
| − | Imagine that we run the computer many times. In roughly <math>1/3</math> of all cases, Bob will be assigned to the same group as Al was. Out of these, in roughly <math>1/3</math> cases Carol will be assigned to the same group as her friends. Thus the probability that all three are in the same group is <math>\frac{1}{3}\cdot \frac{1}{3} = | + | Imagine that we run the computer many times. In roughly <math>1/3</math> of all cases, Bob will be assigned to the same group as Al was. Out of these, in roughly <math>1/3</math> cases Carol will be assigned to the same group as her friends. Thus the probability that all three are in the same group is <math>\frac{1}{3}\cdot \frac{1}{3} = \frac{1}{9}</math>, or <math>\boxed{\text{B}}</math>. |
(The exact value is <math>\frac{199}{599} \cdot \frac{198}{598}</math>, which is <math>\sim 0.11\%</math> less than our approximate answer.) | (The exact value is <math>\frac{199}{599} \cdot \frac{198}{598}</math>, which is <math>\sim 0.11\%</math> less than our approximate answer.) | ||
Revision as of 16:36, 26 January 2009
Problem
The 
students at King Middle School are divided into three groups of equal size for lunch. Each group has lunch at a different time. A computer randomly assigns each student to one of three lunch groups. The probability that three friends, Al, Bob, and Carol, will be assigned to the same lunch group is approximately
Solution
Imagine that we run the computer many times. In roughly 
of all cases, Bob will be assigned to the same group as Al was. Out of these, in roughly cases Carol will be assigned to the same group as her friends. Thus the probability that all three are in the same group is 
, or 
.
(The exact value is 
, which is 
less than our approximate answer.)
