Probability

[匿名]

2010-08-19 6:02 am

In a game, probability that the player wins or loses 1 point at each trial are 3/4 and 1/4 respectively. The game ends when the player's score is +2 or - 2. Find :
(a) Probability that he wins the game.
(b) Probability that he wins the game given that he won the first trial.
(c) Probability that he wins the game given that he lost the first trial.
(d) Probability that he wins the game at the end of the nth trial if (i) n is even, (ii) n is odd.
[Ans. : (a) 9/10 (b) 39/40 (c) 27/40 (di) (9/10)(3/8)^(n/2 - 1) (dii) 0]

回答 (1)

用戶2053

2010-08-19 8:49 am

✔ 最佳答案

a)The last and the 2nd last trial must be WW or LL ,So P( he wins )= (3/4)^2 / [(3/4)^2 + (1/4)^2]= (9/16) / (10/16)= 9/10
b)P(he wins the game given that he won the first trial)= P(W,W) + P(W,L..........W,W)= P(W,W) + P(W,L) * P( he wins )= (1)(3/4) + (1)(1/4) * 9/10= 3/4 + 9/40= 39/40
c)P(he wins the game given that he lost the first trial)= P(L,W,.......W,W)= P(L,W) * P(he wins)= (1)(3/4) * 9/10= 27/40
di)P(he wins the game at the end of the nth trial if n is even)= P[(L,W or W,L) , (L,W or W,L) , (L,W or W,L) , ... W , W].......|.........(n - 2)/2 terms of (L,W or W,L).........|
= { [(1/4)(3/4) + (3/4)(1/4)] ^ [(n - 2)/2] } (3/4)(3/4)
= (9/16)(3/8) ^ (n/2 - 1)(I think your answer (9/10)(3/8)^(n/2 - 1) have typing error.)
ii)It is impossible as you can see that there are even terms in below : [(L,W or W,L) , (L,W or W,L) , (L,W or W,L) , ... W , W] P(E) = 0

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