 #1
jisbon
 476
 30
 Homework Statement:

The number of customers arriving at a shop is on average 1 per weekdays.
The manager is checking customer flow during a 10 min period from 8.00 am
He noted that by 8.05 am, there were at least 3 people walking in.
Find the probability that at most 6 customers walkin within 10 minutes from 8.008.10am
 Relevant Equations:
 
Confused and not sure if it is correct, but please do correct my steps.
We let event B be that there are at least 3 customers entering in 5 minutes.
Hence P(B) = 1 P(X=0) P(X=1)  P(X=2) = ##1 \dfrac{e^{5}5^{0}}{0!}\dfrac{e^{5}5^{1}}{1!}\dfrac{e^{5}5^{2}}{2!} ## = 0.8753...
Now we let event A be there is at most 6 within 10 minutes.
P(A) =P(X=0)+ P(X=1) +P(X=2) ... P(X=6) = ##\:\:\frac{e^{10}10^0}{0!}+\frac{e^{10}10^1}{1!}+\frac{e^{10}10^2}{2!}\:+\frac{e^{10}10^3}{3!}\:+\frac{e^{10}10^4}{4!}\:+\frac{e^{10}10^5}{5!}\:+\:\:\frac{e^{10}10^6}{6!}##=0.13...
Now base on Conditional probability, P(A intersect B)/P(B) will be the probability here, so how do I exactly calculate P (A intersect B)? Is my train of thought correct? Or am I totally wrong here?
Thanks
We let event B be that there are at least 3 customers entering in 5 minutes.
Hence P(B) = 1 P(X=0) P(X=1)  P(X=2) = ##1 \dfrac{e^{5}5^{0}}{0!}\dfrac{e^{5}5^{1}}{1!}\dfrac{e^{5}5^{2}}{2!} ## = 0.8753...
Now we let event A be there is at most 6 within 10 minutes.
P(A) =P(X=0)+ P(X=1) +P(X=2) ... P(X=6) = ##\:\:\frac{e^{10}10^0}{0!}+\frac{e^{10}10^1}{1!}+\frac{e^{10}10^2}{2!}\:+\frac{e^{10}10^3}{3!}\:+\frac{e^{10}10^4}{4!}\:+\frac{e^{10}10^5}{5!}\:+\:\:\frac{e^{10}10^6}{6!}##=0.13...
Now base on Conditional probability, P(A intersect B)/P(B) will be the probability here, so how do I exactly calculate P (A intersect B)? Is my train of thought correct? Or am I totally wrong here?
Thanks