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

(→‎Solution 2)
(rewrite solution as both solutions (esp. solution 2) were lacking strong enough justification, and should mention that the probabilities are approximate.)
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<math>\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}</math>
 
<math>\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}</math>
  
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==Solution 1==
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==Solution==
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There are <math>\binom{3}{1}</math> ways to choose which group the three kids are in and the chance that all three are in the same group is <math>\frac{1}{27}</math>. Hence <math>\frac{1}{9}</math> or <math>\boxed {B}</math>.
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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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==Solution 2==
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We remark that the actual probability is <math>\frac{3 \times \binom{597}{197} \times \binom{400}{200}}{\binom{600}{200} \times \binom{400}{200}} = \frac{3 \times \frac{597!}{400!197!}}{\frac{600!}{400!200!}} = \frac{3 \times 200 \times 199 \times 198}{600 \times 599 \times 598} \approx 0.11</math>.
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One of the statements, that there are <math>600</math> students in the school is redundant. Taking that there are <math>3</math> students and there are <math>3</math> groups, we can easily deduce there are <math>81</math> ways to group the <math>3</math> students, and there are <math>3</math> ways to group them in the same <math>1</math> group, so we might think <math>\frac{3}{81}=\frac{1}{27}</math> is the answer but as there are 3 groups we do <math>\frac{1}{27} (3)=\frac{1}{9}</math> which is <math>\boxed{\text{(B)}}</math>.
 
  
 
==See Also==
 
==See Also==

Revision as of 18:53, 15 March 2023

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 $\boxed{\text{(B)}}$.

We remark that the actual probability is $\frac{3 \times \binom{597}{197} \times \binom{400}{200}}{\binom{600}{200} \times \binom{400}{200}} = \frac{3 \times \frac{597!}{400!197!}}{\frac{600!}{400!200!}} = \frac{3 \times 200 \times 199 \times 198}{600 \times 599 \times 598} \approx 0.11$.

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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