Question #283933

There are 4 traffic lights along the road, each prohibits further movement of the car with a probability of 0.5. Find the

distribution of the number of the traffic lights that car passed to the first stop. What is the variance of the random

variable? 



Expert's answer

Let X=X= the number of the traffic lights that car passed to the first stop.

Then


P(X=0)=0.5(0.5)0=0.5P(X=0)=0.5(0.5)^0=0.5

P(X=1)=0.5(0.5)1=0.25P(X=1)=0.5(0.5)^1=0.25

P(X=2)=0.5(0.5)2=0.125P(X=2)=0.5(0.5)^2=0.125

P(X=3)=0.5(0.5)3=0.0625P(X=3)=0.5(0.5)^3=0.0625

P(X=4)=0.5(0.5)4=0.03125P(X=4)=0.5(0.5)^4=0.03125

x01234p0.50.250.1250.06250.03125\def\arraystretch{1.5} \begin{array}{c:c} x & 0 & 1 & 2 & 3 & 4 \\ \hline p & 0.5 & 0.25 & 0.125 & 0.0625 & 0.03125 \\ \end{array}

E(X)=0(0.5)+1(0.25)+2(0.125)E(X)=0(0.5)+1(0.25)+2(0.125)

+3(0.0625)+4(0.03125)=0.8125+3(0.0625)+4(0.03125)=0.8125

E(X2)=02(0.5)+12(0.25)+22(0.125)E(X^2)=0^2(0.5)+1^2(0.25)+2^2(0.125)

+32(0.0625)+42(0.03125)=1.8125+3^2(0.0625)+4^2(0.03125)=1.8125

Var(X)=E(X2)(E(X))2Var(X)=E(X^2)-(E(X))^2

=1.8125(0.8125)2=1.15234375=1.8125-(0.8125)^2=1.15234375


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