Respuesta :
Answer:
a) The probability that exactly eight arrive during the hour and all eight have no violations is 0.0005.
b) For any fixed y ≥ 8, the probability that y arrive during the hour, of which eight have no violations is:
[tex]P(X=y\,\&\,nv=8)=\frac{4^ye^{-8}}{8!(y-8)!}[/tex]
c) The probability that eight "no-violation" cars arrive during the next hour is 0.030.
Step-by-step explanation:
a) The probability that exactly eight arrive during the hour and all eight have no violations is equal to the product of the probability of arrival of 8 vehicules and the probability of having 8 vehicules with no violations.
[tex]P(X=8\,\&\,no\, violations)=P(no\, violations|X=8)*P(X=8)\\\\P(X=8\,\&\,nv)=(0.5)^8*\frac{8^8e^{-8}}{8!} = 0.0039*0.1396=0.0005[/tex]
b) For any fixed y ≥ 8, the probability that y arrive during the hour, of which eight have no violations is:
[tex]P(X=y\,\&\,nv=8)=P(nv=8|X=y)*P(X=y)\\\\P(X=y\,\&\,nv=8)=[\binom{y}{8}(0.5)^8*(0.5)^{y-8}]*\frac{8^ye^{-8}}{y!} =\frac{y!}{8!(y-8)!}0.5^y *8^y*\frac{e^{-8}}{y!}\\\\ P(X=y\,\&\,nv=8)=(\frac{y!}{y!})(0.5*8)^y\frac{e^{-8}}{8!(y-8)!}=\frac{4^ye^{-8}}{8!(y-8)!}[/tex]
c) Using the result of point (b) we can express the probability that eight "no violation" vehicules arrive durting the next hour as:
[tex]P(nv=8)=\sum\limits^\infty_{y=8} {\frac{4^ye^{-8}}{8!(y-8)!}}=\frac{e^{-8}}{8!} \sum\limits^\infty_{y=8} {\frac{4^y}{(y-8)!}}=\frac{e^{-8}4^{8}}{8!} \sum\limits^\infty_{y=8} {\frac{4^{y-8}}{(y-8)!}}\\\\P(nv=8)=\frac{e^{-8}4^{8}}{8!} \sum\limits^\infty_{z=0} {\frac{4^{z}}{z!}}=\frac{e^{-8}4^{8}}{8!}*e^4=\frac{e^{-4}4^{8}}{8!}\\\\P(nv=8)= 0.030[/tex]
