probability of summation of 4 random variables

Question

i have this question here;

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).

My calculation as such. n = 4, p = 0.2, x <=3 Using binomial formula, the probability distribution is as such
x = 0, 0.4096
x = 1, 0.4096
x = 2, 0.1536
x = 3, 0.0256
x = 4, 0.0016
Adding x = 3 and below, total probability for P(X1 + X2 + X3 + X4 <= 3) is 0.9984

Is 0.9984 correct?

Welcome to Mathematics Stack Exchange! Could you please add an edit to your question explaining how you arrived at the answer you stated / your ideas and attempts at this problem? — Ujan Chakraborty, Jul 18 '21 at 06:11
Under what circumstances, zero, can the sum of the four variables fail to be less than or equal to $3$? — Gerry Myerson, Jul 18 '21 at 07:14

score 1 · Accepted Answer · answered Jul 21 '21 at 01:45

1

Indeed. Alternatively you could use the rule of complementary events.

$$\begin{align}\mathsf P(S\leq 3) &=1-\mathsf P(S=4) \\ &=1-(0.2)^4 \\ &= 0.9984\end{align}$$

Where $S$ is the sum of the four iid random variables.

answered Jul 21 '21 at 01:45

Graham Kemp

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probability of summation of 4 random variables

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