Re: [機統] 基本機率事件-台大電子所
※ 引述《martine318 ()》之銘言：
: In a contest , contestants A , B and C are asked , in turn , a question.
: If a contestant gives a wrong answer to the question , he or she will be
: dropped out of the game. the remaining two will continue to compete
: until one of them drops out. The last person remaining is the winner.
: Suppose that a contestant knows the answer to a question independently
: of the other contestants , with probability P. Let ABC represent the
: event that A drops first , B next , and C wins . Calculate the probability
: of P{ABC)
: 解答 : (1-P)^2 + P^3(1-P)^2+P^6(1-P)^2+......
: +(1-P)^2*P^2+P^3(1-P)^2*P^2+P^6(1-P)^2*P^2+.....
: +(1-P)^2*P^4+P^3(1-P^2)*P^4+P^6(1-P^2)*P^4+.....
應該還要再繼續寫......
: ~
: = (1-P)^2/(1-P^2)*(1/1-P^3)
: 想請問一下列式成上面是如何列的，謝謝!!
如cuttlefish表達的
當A在第k輪答錯 B也同一輪接著答錯的機率 = (P^3)^(k-1) (1 - P)^2
∞
所以AB在同一輪答錯的總機率 = Σ P^(3(k-1)) (1 - P))^2
k=1
當A在第k輪答錯 B也第k+1輪接著答錯的機率 = (P^3)^(k-1) (1 - P) P^2 (1 - P)
∞
所以A比B早一輪答錯的總機率 = Σ P^(3(k-1)) (1 - P))^2 P^2
k=1
當A在第k輪答錯 B也第k+m輪接著答錯的機率 = (P^3)^(k-1) (1 - P) (P^2)^(m) (1 - P)
∞
所以A比B早m輪答錯的總機率 = Σ P^(3(k-1)) (1 - P))^2 P^(2m)
k=1
∞ ∞
最後就是取和 = Σ Σ P^(3(k-1)) (1 - P))^2 P^(2m)
m=0 k=1
= (1 - P)^2 * [1 / (1 - P^2)] * [1 / (1 - P)^3]
而你只寫到m = 3，所以我才說你應該要繼續寫，
至少表達還有類似的列。
解答的做法我覺得不好，
明明可以用很單純的方式解決
當A在第k輪答錯，則B在A之後答錯的機率
= (P^3)^(k - 1) (1 - P)[1 + P^2 + P^4 + ...](1 - P)
= P^(3(k - 1))[1 / (1 - P^2)](1 - P)^2
所以總共的機率
∞
= Σ P^(3(k - 1))[1 / (1 - P^2)](1 - P)^2
k=1
= [1 / (1 - P^3)] * [1 / (1 - P^2)] * (1 - P)^2
--
※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 61.56.10.112
※ 文章網址: https://www.ptt.cc/bbs/Math/M.1482807863.A.E12.html
推
12/27 11:24, , 1F
12/27 11:24, 1F
討論串 (同標題文章)
完整討論串 (本文為第 2 之 2 篇)：