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Difference between revisions of "2002 AIME II Problems/Problem 12"

(changed a10 < .4 to a10 = .4, making the problem actually solvable!)
(Solved!)
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== Solution ==
 
== Solution ==
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{{solution}}
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The first restriction is that <math>a_{10} = .4</math>, meaning that the player gets exactly 4 out of 10 baskets.  The second resetriction is <math>a_n\le.4</math>.  This means that the player may never have a shooting average over 40%.  Thus, the first and second shots must fail, since <math>\frac{1}{1}</math> and <math>\frac{1}{2}</math> are both over <math>.4</math>, but the player may make the third basket, since <math>\frac{1}{3} \le .4</math>  In other words, the earliest the first basket may be made is attempt 3.  Using similar reasoning, the earliest the second basket may be made is attempt 5, the earliest the third basket may be made is attempt 8, and the earliest the fourth basket may be made is attempt 10.
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 +
Using X to represent a basket and O to represent a failure, this 'earliest' solution may be represented as:
 +
 
 +
OOXOXOOXOX
 +
 
 +
To simplify counting, note that the first, second, and tenth shots are predetermined.  The first two shots must fail, and the last shot must succeed.  Thus, only slots 3-9 need to be counted, and can be abbreviated as follows:
 +
 
 +
XOXOOXO
 +
 
 +
The problem may be separated into five cases, since the first shot may be made on attempt 3, 4, 5, 6, or 7.  The easiest way to count the problem is to remember that each X may slide to the right, but NOT to the left.
 +
 
 +
First shot made on attempt 3:
 +
 
 +
XOXOOXO
 +
 
 +
XOXOOOX
 +
 
 +
XOOXOXO
 +
 
 +
XOOXOOX
 +
 
 +
XOOOXXO
 +
 
 +
XOOOXOX
 +
 
 +
XOOOOXX
 +
 
 +
'''Total - 7'''
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 +
First shot made on attempt 4:
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 +
Note that all that needs to be done is change each line in the prior case from starting with "XO....." to "OX.....".
 +
 
 +
'''Total - 7'''
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First shot made on attempt 5:
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 +
OOXXOXO
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 +
OOXXOOX
 +
 
 +
OOXOXXO
 +
 
 +
OOXOXOX
 +
 
 +
OOXOOXX
 +
 
 +
'''Total - 5'''
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First shot made on attempt 6:
 +
 
 +
OOOXXXO
 +
 
 +
OOOXXOX
 +
 
 +
OOOXOXX
 +
 
 +
'''Total - 3'''
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First shot made on attempt 7:
 +
 
 +
OOOOXXX
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'''Total - 1'''
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 +
The total number of ways the player may satisfy the requirements is <math>7+7+5+3+1=23</math>.
 +
 
 +
The chance of hitting any individual combination (say, for example, OOOOOOXXXX) is <math>\left(\frac{3}{5}\right)^6\left(\frac{2}{5}\right)^4</math>
 +
 
 +
Thus, the chance of hitting any of these 23 combinations is <math>23\left(\frac{3}{5}\right)^6\left(\frac{2}{5}\right)^4 = \frac{23\cdot3^6\cdot2^4}{5^{10}}</math>
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 +
Thus, the final answer is <math>(23+3+2+5)(6+4+10)=660</math>
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== See also ==
 
== See also ==
 
{{AIME box|year=2002|n=II|num-b=11|num-a=13}}
 
{{AIME box|year=2002|n=II|num-b=11|num-a=13}}

Revision as of 02:26, 16 December 2008

Problem

A basketball player has a constant probability of $.4$ of making any given shot, independent of previous shots. Let $a_n$ be the ratio of shots made to shots attempted after $n$ shots. The probability that $a_{10} = .4$ and $a_n\le.4$ for all $n$ such that $1\le n\le9$ is given to be $p^aq^br/\left(s^c\right)$ where $p$, $q$, $r$, and $s$ are primes, and $a$, $b$, and $c$ are positive integers. Find $\left(p+q+r+s\right)\left(a+b+c\right)$.

Solution

The first restriction is that $a_{10} = .4$, meaning that the player gets exactly 4 out of 10 baskets. The second resetriction is $a_n\le.4$. This means that the player may never have a shooting average over 40%. Thus, the first and second shots must fail, since $\frac{1}{1}$ and $\frac{1}{2}$ are both over $.4$, but the player may make the third basket, since $\frac{1}{3} \le .4$ In other words, the earliest the first basket may be made is attempt 3. Using similar reasoning, the earliest the second basket may be made is attempt 5, the earliest the third basket may be made is attempt 8, and the earliest the fourth basket may be made is attempt 10.

Using X to represent a basket and O to represent a failure, this 'earliest' solution may be represented as:

OOXOXOOXOX

To simplify counting, note that the first, second, and tenth shots are predetermined. The first two shots must fail, and the last shot must succeed. Thus, only slots 3-9 need to be counted, and can be abbreviated as follows:

XOXOOXO

The problem may be separated into five cases, since the first shot may be made on attempt 3, 4, 5, 6, or 7. The easiest way to count the problem is to remember that each X may slide to the right, but NOT to the left.

First shot made on attempt 3:

XOXOOXO

XOXOOOX

XOOXOXO

XOOXOOX

XOOOXXO

XOOOXOX

XOOOOXX

Total - 7

First shot made on attempt 4:

Note that all that needs to be done is change each line in the prior case from starting with "XO....." to "OX.....".

Total - 7

First shot made on attempt 5:

OOXXOXO

OOXXOOX

OOXOXXO

OOXOXOX

OOXOOXX

Total - 5

First shot made on attempt 6:

OOOXXXO

OOOXXOX

OOOXOXX

Total - 3

First shot made on attempt 7:

OOOOXXX

Total - 1

The total number of ways the player may satisfy the requirements is $7+7+5+3+1=23$.

The chance of hitting any individual combination (say, for example, OOOOOOXXXX) is $\left(\frac{3}{5}\right)^6\left(\frac{2}{5}\right)^4$

Thus, the chance of hitting any of these 23 combinations is $23\left(\frac{3}{5}\right)^6\left(\frac{2}{5}\right)^4 = \frac{23\cdot3^6\cdot2^4}{5^{10}}$

Thus, the final answer is $(23+3+2+5)(6+4+10)=660$







See also

2002 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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