Answer:
a) X1+X2 have distribution Bi(n1+n2, p)
b)
P(X1+X2 = 1 | X2 = 0) = np(1-p)ⁿ¹⁻¹
P(X1+X2 = 1| X2 = 1) = (1-p)ⁿ¹
P(X1 + X2 = 1) = (1-p)ⁿ¹ * np(1-p)ⁿ²⁻¹+ (1-p)ⁿ²*np(1-p)ⁿ¹-¹
Step-by-step explanation:
Since both variables are independent but they have the same probability parameter, you can interpret that like if the experiment that models each try in both variables is the same. When you sum both random variables toguether, what you obtain as a result is the total amount of success in n1+n2 tries of the same experiment, thus X1+X2 have distribution Bi(n1+n2, p).
b)
Note that, if X2 = k, then X1+X2 = 1 is equivalent to X1 = 1-k. Since X1 and X2 are independent, then P(X1+X2 = 1| X2 = K) = P(X1=1-k|X2=k) = P(X1 = 1-k).
If k = 0, then this probability is equal to P(X1 = 1) = np(1-p)ⁿ¹⁻¹
If k = 1, then it is equal to P(X1 = 0) = (1-p)ⁿ¹
Thus,
P(X1+X2 = 1) = P(X1+X2 = 1| X2 = 1) * P(X2=1) + P(X1+X2 = 1| X2 = 0) * P(X2 = 0) = (1-p)ⁿ¹ * np(1-p)ⁿ²⁻¹+ (1-p)ⁿ²*np(1-p)ⁿ¹-¹
