More about probability

Amy and Bob play a game with each other. The probabilities of Amy winning, Bob winning and a tie are 1/2,1/3 and 1/6 respectively. If Amy and Bob play the game 3 times, find the probabilities that Bob wins three times, Bob wins at least one, Amy and Bob win alternately.

A = Amy (A) winning
B = Bob (B) winning
T = Tie

P(A) = 1/2
P(B) = 1/3
P(T) = 1/6

P(Bob wins three times)
= P(BBB)
= [P(B)]^3
= (1/2)^3
= 1/8

P(Bob wins at least one)
= 1 - P(Bob loses all three games)
= 1 - [P(A or T)]^3
= 1 - [(1/2) + (1/6)]^3
= 1 - (2/3)^3
= 1 - (8/27)
= 19/27

P(Amy and Bob win alternately)
= P(ABA or BAB)
= (1/2)*(1/3)*(1/2) + (1/3)*(1/2)*(1/3)
= (1/12) + (1/18)
= 5/36


= = = = =
Harry is going to take part in 50 m breast stroke,100m free style and 200m relay in the swimming gala. It is known that his probabilities of winning the above three races are 1/2, 3/4 and x respectively, If the probability that Harry wins at least one race is 9/10,find x.
(The second probability should be 3/4 instead of 4/4, and it is amended above.)

P(Harry wins at least one race) = 1- P(Harry loses all the three games)
9/10 = 1 - [1 - (1/2)]*[1 - (3/4)]*(1 - x)
1 - (1/8)*(1 - x) = 9/10
(1/8)*(1 - x) = 1/10
1 - x = 4/5
x = 1/5

For Q2, is the second probability = 4/4 ?

2011-04-19 20:40:08 補充：
yes, that's why I need to confirm that if there's any typo in the Q
Cos if the second prob. = 4/4 = 1, then Harry is for sure to win at least one race