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} = \boxed{\frac{1}{9}}</math>.
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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 17:36, 26 January 2009

Problem

The $600$ 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

$\text{(A)}\ \frac{1}{27} \qquad \text{(B)}\ \frac{1}{9} \qquad \text{(C)}\ \frac{1}{8} \qquad \text{(D)}\ \frac{1}{6} \qquad \text{(E)}\ \frac{1}{3}$

Solution

Imagine that we run the computer many times. In roughly $1/3$ of all cases, Bob will be assigned to the same group as Al was. Out of these, in roughly $1/3$ cases Carol will be assigned to the same group as her friends. Thus the probability that all three are in the same group is $\frac{1}{3}\cdot \frac{1}{3} = \frac{1}{9}$, or $\boxed{\text{B}}$.

(The exact value is $\frac{199}{599} \cdot \frac{198}{598}$, which is $\sim 0.11\%$ less than our approximate answer.)

See Also

1986 AJHSME Problems

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