- #1

jisbon

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- 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 walk-in within 10 minutes from 8.00-8.10am

- Relevant Equations
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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