Answer to Question #218449 in Statistics and Probability for Hanan Rajpoot

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

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

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

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