Difference between revisions of "1989 AHSME Problems/Problem 30"
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<math>\text{(A)}\ 9\qquad\text{(B)}\ 10\qquad\text{(C)}\ 11\qquad\text{(D)}\ 12\qquad\text{(E)}\ 13</math> | <math>\text{(A)}\ 9\qquad\text{(B)}\ 10\qquad\text{(C)}\ 11\qquad\text{(D)}\ 12\qquad\text{(E)}\ 13</math> | ||
| − | ==Solution== | + | ==Solution 1== |
| + | |||
| + | We approach this problem using Linearity of Expectation. Consider a pair of two people standing next to each other. Ignoring all other people, the probability that a boy is standing on the left position and a girl is standing on the right position is <math>\frac7{20}\cdot\frac{13}{19}</math>. Similarly, if a girl is standing on the left position and a boy is standing on the right position the probability is also <math>\frac{7\cdot 13}{20\cdot 19}</math>. Thus, the total probability of the two people being one boy and one girl is <math>\frac{91}{190}</math>.\\ \\ | ||
| + | There are a total of 19 different adjacent pairs, so by Linearity of Expectation, we have that the expected value of <math>S</math> is <math>\frac{91}{10} \to \boxed{A}</math>. | ||
| + | |||
| + | ==Solution 2== | ||
Suppose that the class tried every configuration. Boy <math>i</math> and girl <math>j</math> would stand next to each other in <math>2</math> different orders, in <math>19</math> different positions, <math>18!</math> times each. Summing over all <math>i,j</math> gives <math>7\cdot13\cdot2\cdot19\cdot18!=\tfrac{91}{10}\cdot20!</math>, so the average value of <math>S</math> is <math>\boxed{9\tfrac1{10}(A)}</math>. | Suppose that the class tried every configuration. Boy <math>i</math> and girl <math>j</math> would stand next to each other in <math>2</math> different orders, in <math>19</math> different positions, <math>18!</math> times each. Summing over all <math>i,j</math> gives <math>7\cdot13\cdot2\cdot19\cdot18!=\tfrac{91}{10}\cdot20!</math>, so the average value of <math>S</math> is <math>\boxed{9\tfrac1{10}(A)}</math>. | ||
Revision as of 16:13, 24 August 2020
Contents
Problem
Suppose that 7 boys and 13 girls line up in a row. Let 
be the number of places in the row where a boy and a girl are standing next to each other. For example, for the row 
we have that 
. The average value of (if all possible orders of these 20 people are considered) is closest to
Solution 1
We approach this problem using Linearity of Expectation. Consider a pair of two people standing next to each other. Ignoring all other people, the probability that a boy is standing on the left position and a girl is standing on the right position is 
. Similarly, if a girl is standing on the left position and a boy is standing on the right position the probability is also 
. Thus, the total probability of the two people being one boy and one girl is 
.\\ \\
There are a total of 19 different adjacent pairs, so by Linearity of Expectation, we have that the expected value of is 
.
Solution 2
Suppose that the class tried every configuration. Boy 
and girl 
would stand next to each other in 
different orders, in 
different positions, 
times each. Summing over all 
gives 
, so the average value of is 
.
See also
| 1989 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 29 |
Followed by Last Question | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
