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>. | ||
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== See also == | == See also == |
Latest revision as of 17:13, 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 |
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