Probability

With reference to the competition diagram below:


圖片參考：http://i388.photobucket.com/albums/oo325/loyitak1990/Jul10/Crazyprob1.jpg


Since the players are randomly assigned, the probability of each player's assignment to positions 1 - 8 are equal.

Also, suppose that the following events are considered:

E1: A vs B in quarter-final

E2: A wins the champion

Then:

a) P(E1) = 1 x 1/7 = 1/7

since when the first player is drawn, he/she can be assigned to any position and then for the second player, he/she should be assigned to the same match with the first player in quarter final, i.e. one acceptable position out of 7 possibilities.

b) P(E2) = 1/2 x 1/2 x 1/2 = 1/8

since A has to win 3 matches consecutively to win the champion.

Thus the required probability for this part is:

P(E1 and E2 | E2) = (1/7 x 1/2 x 1/2 x 1/2)/(1/8) = 1/7

2010-07-13 14:33:03 補充：
Image URL:
http://i388.photobucket.com/albums/oo325/loyitak1990/Jul10/Crazyprob1.jpg

2010-07-13 17:02:53 補充：
In Chinese, quarter final is 八強, semi final is 四強

a) P(E)

= P(A not vs B) * P(Both A and B win the 1st round) * P(Then A vs B in the quarter-final)

= (6/7) * (1/2)(1/2) * (1/3)

= 1/14

2010-07-13 16:45:46 補充：
b)P(E)

= (6/7) * (1)(1/2) * (1/3)

= 1/7

May be wrong !!

2010-07-13 21:25:03 補充：
quarter final is 八強, semi final is 四強

Understand.

I wrong.

1 5
---- A C ---
2 6
----A------A---- C------
3 7
--- B D ---
4 8


a) 1/2x1/3x1/2

= 1/12