Three cases:
only ann wins : (1/3)(7/8)(19/20) = 13/480
only may wins: (2/3)(1/8)(19/20) = 38/480
only vea wins : (2/3)(7/8)(1/20) = 14/480
prob(as stated) = 13/480 + 38/480 + 14/480
= ....
If ann, may, and vea had 1/3, 1/8, and 1/20 chances of winning a competition, what is the probability that only one of them will win?
*I don't know if I would just add everything and that would be the amswer, or is there something more?*
1 answer
1 answer