Question

Let X1, X2, X3, X4 be random variables that are all independent of each other and have the same distribution, namely, P(X1 = 1) = 0.2, P(X1 = 0) = 0.8, and identically so for X2, X3, X4. Calculate the probability that P(X1 + X2 + X3 + X4 <= 3)?

Answer 1

$$P(X_1+X_2+X_3+X_4\leq 3)$$

$$=1-P(X_1+X_2+X_3+X_4=4)$$

$$=1-0.2(0.2)(0.2)(0.2)=0.9984$$