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

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==Solution 2==
 
==Solution 2==
 
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}{54}=\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>.
 
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}{54}=\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>.
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==Solution 3 (easiest)==
 
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The information that there are <math>600</math> students is irrelevant. The first student has <math>3</math> choices to choose from. In order for all students to go in the same lunch group, the second student has <math>1</math> choice, and same for the third student.
 
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~sakshamsethi
 
  
 
==See Also==
 
==See Also==

Revision as of 11:02, 28 July 2020

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 1

There are $\binom{3}{1}$ ways to choose which group the three kids are in and the chance that all three are in the same group is $\frac{1}{27}$. Hence $\frac{1}{9}$ or $\boxed {B}$.

Solution 2

One of the statements, that there are $600$ students in the school is redundant. Taking that there are $3$ students and there are $3$ groups, we can easily deduce there are $81$ ways to group the $3$ students, and there are $3$ ways to group them in the same $1$ group, so we might think $\frac{3}{54}=\frac{1}{27}$ is the answer but as there are 3 groups we do $\frac{1}{27} (3)=\frac{1}{9}$ which is $\boxed{\text{(B)}}$.

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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