A person plays m independent games.The probability of him winning any game is a/(a+b) (a,b are positive numbers).show that probability that the person wins odd number of games is
1/2*[(b+a)^m - (b-a)^m]/(b+a)^m

Question

A person plays m independent games.The probability of him winning any game is a/(a+b) (a,b are positive numbers).show that probability that the person wins odd number of games is 
1/2*[(b+a)^m - (b-a)^m]/(b+a)^m

anonymous · Answer

I need a clear instruction of the prove

shubhamsrg · Answer

What will be the probability of winning only 1 game ?

shubhamsrg · Answer

out of m

shubhamsrg · Answer

@Seran

anonymous · Answer

mC1*(a/(a+b))^1*(1-a/(a+b))^m-1

shubhamsrg · Answer

correct
same logic goes for winning 3 games
and so on..

there will be two cases :
1) m is odd
2) m-1 is odd

anonymous · Answer

so i need to find both cases and add their prob isn't it

anonymous · Answer

I mean for m is odd sum of all odds and for m-1 is odd sum of all odds and add them both is it

shubhamsrg · Answer

I think it'll be : P(m is odd)*P(odd no. of wins) + P(m is even) * P(odd no. of wins)
P(m is odd)= 1/2
See if this matches your ans ?