A game has 9 trials. In any trial, three possibilities can be – win, lose, and draw. What will be the probability for at least one win and one draw?

Question

tkhunny · Answer

Insufficient information.  Are the outcomes equally likely?

Calculate this: $(W + L + D)^{3}$.  It won't take too long.

anonymous · Answer

are u saying how many wins and lose and draw like 3 wins 2 draw,4lose like wise?

tkhunny · Answer

I'm saying there is insufficient information.

If you REALLY want to answer the question in SOME way, you'll have to make some assumptions.   One possible assumption is that the outcomes are equally likely:

p(Win) = p(Loss) = p(Draw) = 1/3

There are infinitely many other assumptions that could be made.

anonymous · Answer

yes .ur correct! p(Win) = p(Loss) = p(Draw) = 1/3. this is the assumption

anonymous · Answer

but if u assume we need these data that how many wins los e r draw

anonymous · Answer

count the compliment 
no win or no draw = (no win) + (no draw) - (no win and no draw)
= 2^9 + 2^9 - 1

3^3 - (2^9 + 2^9 - 1) = 18660

P = 18660/3^9 = 6220/6561 <--