Hint the hypergeometric distribution may be helpful in findi

(Hint: the hypergeometric distribution may be helpful in finding the final marginal p.m.f.)

Solution

[ the solution is given using MGF transformation]

X1~Bin(n1,p) X2~Bin(n2,p) X1 and X2 are independent

then MGF of X1 is M1(t)=(q+pe^t)ⁿ¹          where q=1-p   t be a real number

then MGF of X2 is M2(t)=(q+pe^t)ⁿ²  

let Z=X1+X2

then MGF of Z is M(t)=M1(t)*M2(t) [as X1 and X2 are independent]

                               =(q+pe^t)ⁿ¹⁺ⁿ²

but this is the mgf of a binomial distribution with parameters n1+n2 and p

and as MGF uniquely characterises a distribution hence X1+X2~Bin(n1+n2,p) [proved]

intuitive argument:

here X1 denotes the number of successes in n1 trials with probability of success p

and X2 denotes the number of successed in n2 trials with probability of success p

hence X1+X2 would be the total number of successes in n1+n2 trials

hence X1+X2 must follow Bin(n1+n2,p)