2002 AIME II 第 12 题

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12.

一名篮球运动员每次投篮命中的概率恒为 0.40.4,并且与之前的投篮无关。令 ana_n 为投完 nn 次后命中数与出手数之比。事件 a10=0.4a_{10} = 0.4an0.4a_n \le 0.4 同时发生的概率(后一条件中的 nn 取遍满足 1n91 \le n \le 9 的整数)为 paqbr(sc)\frac{p^a q^b r}{\left(s^c\right)},其中 ppqqrrss 都是素数,aabbcc 都是正整数。求 (p+q+r+s)(a+b+c)(p + q + r + s)(a + b + c)

A basketball player has a constant probability of 0.40.4 of making any given shot, independent of previous shots. Let ana_n be the ratio of shots made to shots attempted after nn shots. The probability that a10=0.4a_{10} = 0.4 and an0.4a_n \le 0.4 for all nn such that 1n91 \le n \le 9 is given to be paqbr(sc),\frac{p^a q^b r}{\left(s^c\right)}, where p,p, q,q, r,r, and ss are primes, and a,a, b,b, and cc are positive integers. Find (p+q+r+s)(a+b+c).(p + q + r + s)(a + b + c).

答案:660
知识点:格路递推计数二项概率
难度评级:2990
小提示:

把(出手次数,命中次数)看作格路径;条件 an0.4a_n \le 0.4 将每次投篮后的命中数限制在 0.4n\lfloor 0.4 n \rfloor 以内。

Track (shots attempted, shots made) as a lattice path; the condition an0.4a_n \le 0.4 caps the number made at 0.4n\lfloor 0.4 n \rfloor after each shot

大提示:

通过把每个点的两个前驱点计数相加,数出到 (10,4)(10, 4) 的受限路径;每个有效序列的概率都是 (0.4)4(0.6)6(0.4)^4 (0.6)^6

Count the constrained paths to (10,4)(10, 4) by adding the counts of the two predecessors at each point; every valid sequence has probability (0.4)4(0.6)6(0.4)^4 (0.6)^6

解答:

把球员的过程记录为点 (n,y)(n, y) 的路径,其中 yynn 次出手后的命中数。条件 an0.4a_n \le 0.4yy 限制在 0.4n\lfloor 0.4n \rfloor 以内;当 n=1,,9n = 1, \ldots, 9 时,这些上限分别为 000011112222223333,而 a10=0.4a_{10} = 0.4 表示路径终点为 (10,4)(10, 4)

在每个允许点处,把它的两个前驱点的计数相加(未命中时 yy 不变,命中时它增加 11)。当 n=3,,9n = 3, \ldots, 9 时,最高允许高度处的计数依次为 1122225599992323;第十次投篮必须命中,所以共有 2323 个投篮序列满足条件。每个序列有 44 次命中和 66 次未中,因此概率为 23(25)4(35)6=24362351023 \left(\tfrac{2}{5}\right)^4 \left(\tfrac{3}{5}\right)^6 = \frac{2^4 \, 3^6 \cdot 23}{5^{10}}\text{。}

因此 {p,q,r,s}={2,3,23,5}\{p, q, r, s\} = \{2, 3, 23, 5\},且 (a,b,c)=(4,6,10)(a, b, c) = (4, 6, 10),得到 (2+3+23+5)(4+6+10)(2 + 3 + 23 + 5)(4 + 6 + 10) =3320= 33 \cdot 20 =660= 660

Record the player’s progress as a path through points (n,y),(n, y), where yy is the number of shots made after nn attempts. The condition an0.4a_n \le 0.4 caps yy at 0.4n,\lfloor 0.4n \rfloor, which for n=1,,9n = 1, \ldots, 9 is 0,0, 0,0, 1,1, 1,1, 2,2, 2,2, 2,2, 3,3, 3,3, and a10=0.4a_{10} = 0.4 means the path ends at (10,4).(10, 4).

Count the allowed paths by adding, at each point, the counts of its two predecessors (a miss keeps y,y, a make raises it by 11). The counts at the maximum allowed heights for n=3,,9n = 3, \ldots, 9 come out to 1,1, 2,2, 2,2, 5,5, 9,9, 9,9, 23,23, and the tenth shot must be a make, so 2323 shot sequences qualify. Each consists of 44 makes and 66 misses, so the probability is 23(25)4(35)6=243623510.23 \left(\tfrac{2}{5}\right)^4 \left(\tfrac{3}{5}\right)^6 = \frac{2^4 \, 3^6 \cdot 23}{5^{10}}.

Thus {p,q,r,s}={2,3,23,5}\{p, q, r, s\} = \{2, 3, 23, 5\} and (a,b,c)=(4,6,10),(a, b, c) = (4, 6, 10), giving (2+3+23+5)(4+6+10)(2 + 3 + 23 + 5)(4 + 6 + 10) =3320= 33 \cdot 20 =660.= 660.

第 11 题#11
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