2002 AIME II Problems/Problem 12
A basketball player has a constant probability of of making any given shot, independent of previous shots. Let be the ratio of shots made to shots attempted after shots. The probability that and for all such that is given to be where , , , and are primes, and , , and are positive integers. Find .
The first restriction is that , meaning that the player gets exactly 4 out of 10 baskets. The second resetriction is . This means that the player may never have a shooting average over 40%. Thus, the first and second shots must fail, since and are both over , but the player may make the third basket, since 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 .
The chance of hitting any individual combination (say, for example, OOOOOOXXXX) is
Thus, the chance of hitting any of these 23 combinations is
Thus, the final answer is
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