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Difference between revisions of "1986 AJHSME Problems/Problem 24"

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==Solution==
 
==Solution==
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Let us first assign Al to a group. We want to estimate the probability that Bob and Carol are assigned to the same group as Al. As the groups are large and of equal size, we can estimate that Bob and Carol each have a <math>\approx \frac{1}{3}</math> probability of being assigned to the same group as Al, and that these events are mostly independent of each other. The probability that all three are in the same lunch group is approximately <math>\left(\frac{1}{3}\right)^2 = \frac{1}{9}</math>, or <math>\boxed{\text{(B)}}</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>.
 
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(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.)
 
  
 
==See Also==
 
==See Also==
  
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[[1986 AJHSME Problems]]
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{{AJHSME box|year=1986|num-b=23|num-a=25}}
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[[Category:Introductory Probability Problems]]
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[[Category:Introductory Combinatorics Problems]]
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{{MAA Notice}}

Latest revision as of 13:07, 26 June 2024

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

Let us first assign Al to a group. We want to estimate the probability that Bob and Carol are assigned to the same group as Al. As the groups are large and of equal size, we can estimate that Bob and Carol each have a $\approx \frac{1}{3}$ probability of being assigned to the same group as Al, and that these events are mostly independent of each other. The probability that all three are in the same lunch group is approximately $\left(\frac{1}{3}\right)^2 = \frac{1}{9}$, or $\boxed{\text{(B)}}$.

See Also

1986 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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