Answer:
0.3968
Step-by-step explanation:
Lets first calculate the probability of the individual to be server in less than 3 minutes. We will call X the amount of time the person had to wait.
[tex]P_X(3) = 1/4* \int\limits^3_0 {e^{-0.25t}} \, dt = 1/4*(-4)*(e^{-0.25t} \, |_0^3) = -1(e^{-0.75}-1) = 1-e^{-0.75} = 0.5276[/tex]
We make this experiment, that has 0.5276 probability of success, 6 times. The total amount of success is a binomial distribution B(6,0.5276). Lets call this random variable Y. Note that
[tex]P(Y \geq 4) = P(Y=4) + P(Y=5)+P(Y=6)[/tex]
[tex] P_Y(4) = {6 \choose 4} * 0.5276^4 * (1-0.5276)^2 = 0.2594[/tex]
[tex] P_Y(5) = {6 \choose 5} *0.5276^5*(1-0.5276) = 0.1158[/tex]
[tex]P_Y(6) = 0.5276^6 = 0.0216 [/tex]
Therefore,
[tex]P(Y \geq 4) = P(Y=4) + P(Y=5)+P(Y=6) = 0.2594+ 0.1158+0.0216 = 0.3968[/tex]
