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

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Thus, the final answer is <math>(23+3+2+5)(6+4+10)=660</math>
Thus, the final answer is <math>(23+3+2+5)(6+4+10)=660</math>
== 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 14:19, 17 December 2008


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 $\left(p+q+r+s\right)\left(a+b+c\right)$.


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:


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:


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:








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:






Total - 5

First shot made on attempt 6:




Total - 3

First shot made on attempt 7:


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 (ProblemsAnswer KeyResources)
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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