m balls are distributed among a boys and b girls. Prove that the probability that odd numbers of balls are distributed to boys is `((b+a)^m - (b+a)^m)/ (2 (a+b)^m)`.

Explanation

Suppose than p and q, respectively, denote the probabilities that a thing goes to a man and a women.
`p=(a)/(a+b)and q=(b)/(a+b)`
Now, the probabilities of 0,1,2,3,… thing going to men are the first, second third terms etc. in the following binomial expansion. `(q+p)^(m)=q^(m)+""^(m)C_(1)a^(m-2)p^(2)+...+p^(m)" "(1)`
But men aer to receive and odd number of things.
Hence the required probability is the sum of even terms in the equation (1).
To obtain the sum of evetn terms, we write the expansion
`(q-p)^(m)=q^(m)-""^(m)C_(1)a^(m-1)p+""^(m)C_(2)a^(m-2)p^(2)-...+(-1)^(m)p^(m)" "(2)`
Subtracting (2) from (1), we get
`(q+p)^(m)-(q-p)^(m)=` Sum of even terms in (1)
Hence, the required probability
`=1/2[(q+p)^(m)-(q-p)^(m)]`
`=1/2[1-((b-a)/(b+a))^(m)]`
`=((b+a)^(m)-(b-a)^(m))/(2(b+a)^(m))`