AIME 2010 II · 第 4 题

Solutions

Solution 1

There are $12 \cdot 11 = 132$ possible situations ($12$ choices for the initially assigned gate, and $11$ choices for which gate Dave's flight was changed to). We are to count the situations in which the two gates are at most $400$ feet apart.

If we number the gates $1$ through $12$, then gates $1$ and $12$ have four other gates within $400$ feet, gates $2$ and $11$ have five, gates $3$ and $10$ have six, gates $4$ and $9$ have have seven, and gates $5$, $6$, $7$, $8$ have eight. Therefore, the number of valid gate assignments is

$$2\cdot(4+5+6+7)+4\cdot8 = 2 \cdot 22 + 4 \cdot 8 = 76$$ so the probability is $\frac{76}{132} = \frac{19}{33}$. The answer is $19 + 33 = \boxed{052}$.

Solution 2

As before, derive that there are $132$ possibilities for Dave's original and replacement gates.

Now suppose that Dave has to walk $100k$ feet to get to his new gate. This means that Dave's old and new gates must be $k$ gates apart. (For example, a $100$ foot walk would consist of the two gates being adjacent to each other.) There are $12-k$ ways to pick two gates which are $k$ gates apart, and $2$ possibilities for gate assignments, for a total of $2(12-k)$ possible assignments for each $k$.

As a result, the total number of valid gate arrangements is

$$2\cdot 11 + 2\cdot 10 + 2\cdot 9 + 2\cdot 8 = 76$$ and so the requested probability is $\tfrac{19}{33}$ for a final answer of $\boxed{052}$.

Solution 3

We can just manually count via each probability.

Note that the probability he gets assigned to gate $1$ is the same as the probability he gets assigned to gate $12$. Mappings are made symmetrically. Thus we find the cases until gate $6$, and multiply each by $2$.

If he gets gate $1$ with chance $\frac{1}{12}$, he can move to gate $2, 3, 4,$ or $5$ with probability $\frac{4}{12}$ after the change. The total probability is $\frac{1}{12} \times \frac{4}{11}$.

If he gets gate $2$ with chance $\frac{1}{12}$, he can move to gate $1, 3, 4, 5,$ or $6$, with combined probability $\frac{5}{12}$. The total probability is $\frac{1}{12} \times \frac{5}{11}$.

If he gets gate $3$ with chance $\frac{1}{12}$, he can move to $1, 2, 4, 5, 6, 7$ with combined probability $\frac{6}{12}$. The total probability is $\frac{1}{12} \times \frac{6}{11}$.

If he gets gate $4$ with chance $\frac{1}{12}$, he can move to $1, 2, 3, 5, 6, 7, 8$ with combined probability $\frac{7}{12}$. The total probability is $\frac{1}{12} \times \frac{7}{11}$.

If he gets gate $5$ with chance $\frac{1}{12}$, he can move to $1, 2, 3, 4, 6, 7, 8, 9$ with combined probability $\frac{8}{12}$. The total probability is $\frac{1}{12} \times \frac{8}{11}$.

Finally, if he gets gate $6$ with chance $\frac{1}{12}$, he can move to $2, 3, 4, 5, 7, 8, 9, 10$ with combined probability $\frac{8}{12}$. The total probability is $\frac{1}{12} \times \frac{8}{11}$.

Then, the combined probability for $1 \text{--} 6$ is:

$$\frac{1}{12} \times \frac{4}{11} + \frac{1}{12} \times \frac{5}{121} + \frac{1}{12} \times \frac{6}{11} + \frac{1}{12} \times \frac{7}{11} + \frac{1}{12} \times \frac{8}{11} + \frac{1}{12} \times \frac{8}{11}$$ or $\frac{1}{12} \left( \frac{4}{11} + \frac{5}{11} + \frac{6}{11} + \frac{7}{11} + \frac{8}{11} + \frac{8}{11} \right)$, or $\frac{1}{132} (4 + 5 + 6 + 7 + 8 + 8)$, or $\frac{1}{132} (38)$ or $\frac{38}{132}$, or $\frac{19}{66}$.

Then, when we multiply the total by $2$, we obtain $\frac{19}{33}$. $19+33 = \boxed{052}$

~Pinotation