Seems Simple Yet Hard

[piecrust123]

Louis, Otis, and Ricky are sitting on the their Grandmas front porch. Cars pass by at a rate of .8 cars per hour (in the boondocks im guessing). If they sit on the porch for 2.5 hours, which brother has the best chance at winning their five dollar bet. Louis thinks that 3 cars will pass by exactly. Otis bets that less than 3 cars will pass by before they get up. Ricky believes they both are wrong.

Let X be the number of cars that pass by while they sit on the porch

From this I have

P(Louis winning bet)=P(X=3)=
P(Otis winning bet)=P(X<3)=
P(Ricky winning bet)=P(X>3)=

Am i right so far and where do i go from here????

 

[Deleted member 4993]

Louis, Otis, and Ricky are sitting on the their Grandmas front porch. Cars pass by at a rate of .8 cars per hour (in the boondocks im guessing). If they sit on the porch for 2.5 hours, which brother has the best chance at winning their five dollar bet. Louis thinks that 3 cars will pass by exactly. Otis bets that less than 3 cars will pass by before they get up. Ricky believes they both are wrong.

Let X be the number of cars that pass by while they sit on the porch

From this I have

P(Louis winning bet)=P(X=3)=
P(Otis winning bet)=P(X<3)=
P(Ricky winning bet)=P(X>3)=

Am i right so far and where do i go from here????

Have you studied expected values?

 

[galactus]

This is a Poisson probability.

\(\displaystyle {\lambda}=2.5(.8)=2\)

Exactly 3:

\(\displaystyle \frac{2^{3}e^{-2}}{3!}\)

Less than 3:

\(\displaystyle \sum_{k=0}^{2}\frac{2^{k}e^{-2}}{k!}\)

In order for them to both be wrong, more than 3 cars will have to drive by:

More than 3:

\(\displaystyle 1-\sum_{k=0}^{3}\frac{2^{k}e^{-2}}{k!}\)

Which one has the higher probability?.