Conditional Probability + Poisson Distribution

[jisbon]

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

 

[kuruman]

As I understand the Poisson distribution, events are randomly occurring, the most common example being radioactive decay. I don't believe that the probability of no more than 6 walk in during a 10 min. period will be any different if 5 as opposed to 3 customers walk in in the first 5 min. period.

Having said that, what is the average value that you use in the Poisson distribution? Is it really 1 per day as the statement of the problem seems to imply or is there something missing? It is seems you are using 10 for the average. Where did that come from?

 

[jisbon]

As I understand the Poisson distribution, events are randomly occurring, the most common example being radioactive decay. I don't believe that the probability of no more than 6 walk in during a 10 min. period will be any different if 5 as opposed to 3 customers walk in in the first 5 min. period.

Having said that, what is the average value that you use in the Poisson distribution? Is it really 1 per day as the statement of the problem seems to imply or is there something missing? It is seems you are using 10 for the average. Where did that come from?

It is 1 per minute per day, sorry for not including that in. Hence in 10 minutes it will be 10 for average

 

[kuruman]

That makes more sense, thank you. I think the answer is P = 0.13 as you calculated. These are random events and there is nothing conditional about them.

 

[jisbon]

That makes more sense, thank you. I think the answer is P = 0.13 as you calculated. These are random events and there is nothing conditional about them.

Oh so its simply the probability that at most 6 customers walk-in within 10 minutes, there's no need to consider the first 5 minutes that at least some customers has walked in?

 

[kuruman]

Yes, it's the cumulative Poisson probability $$P_{cum}=\sum_{k=0}^{6}P(10,k).$$

 

[haruspex]

I don't believe that the probability of no more than 6 walk in during a 10 min. period will be any different if 5 as opposed to 3 customers walk in in the first 5 min. period.

That would be true if it were the next 10 minute period, but the 5 minute and 10 minute periods overlap. I.e. we have at least 3 already, and don't want to exceed 6 total.
(The wording is awkward, but I think it is all referring to the same morning.)

 

[scottdave]

I'm just jotting some ideas down, to see how they look. I'm on my phone right now.

Poisson distributed arrivals with parameter Lambda will have interarrival times with an exponential (Lambda) distributtion.

It asks for "at most" 6, so could we say we want at most 3, during the second 5-minute time period. Hmm. Maybe that's not the approach.

Your parameter for Poisson is for 1 time period (minute), but you need 5 time periods. It could be 3 in 1 minute and none in the next 4.

I'll have to think about the analytical approach, but in the meantime, I'd run a simulation to see what that says. It wouldn't be difficult in R or Python. With enough random trials, your analytical solution should be close to the simulation result.

 

[scottdave]

Oh wait. Poisson rate parameter is how many per Time Period. So what if you say "what is the rate per (5-minute) time period"? I think that might be the approach. The ones in the first five minutes matter.

Here's why. Consider it said that 10 had arrived between 8 and 8:05. What is the probability that "at most" 6 arrived between 8 and 8:10. See?

 

[kuruman]

That would be true if it were the next 10 minute period, but the 5 minute and 10 minute periods overlap. I.e. we have at least 3 already, and don't want to exceed 6 total.
(The wording is awkward, but I think it is all referring to the same morning.)

Yes, it is the same morning but I think the walk-in probability is completely random and therefore obeys the Poisson distribution as the OP has ascertained. Its randomness can be modeled after radioactive decay which certainly obeys the Poisson distribution.

Suppose you have a throng of customers waiting outside the store. Also suppose that the store manager has a sample of unstable and undecayed nuclei with a half-life of centuries and an average counting rate of 1 per minute. The manager has associated one customer's name to each undecayed nucleus. When a particular nucleus decays, the associated customer is allowed to walk in. When or who walks in is completely random.

Once the counting interval of 10-min is specified and the average count over that interval is known, the probabilities of any specific number of counts within that interval are fixed by the Poisson distribution.

On edit:
After some more thought it occurred to me that one may be asked to calculate the probability that at least 3 walk in over the second 5 min interval assuming an average of 5 per 5-min interval. If that's the case, I agree with @haruspex that the wording is awkward. It would have been better to ask "Find the probability that at most 3 customers walk in from 8:05 to 8:10."

 

[haruspex]

After some more thought it occurred to me that one may be asked to calculate the probability that at least 3 walk in over the second 5 min interval assuming an average of 5 per 5-min interval. If that's the case, I agree with @haruspex that the wording is awkward. It would have been better to ask "Find the probability that at most 3 customers walk in from 8:05 to 8:10."

No, that's a different question.
At 8:05 there may already be 3, 4, 5, ... people have visited the shop. We need the probabilities of those, and for each the probability that we haven't reached 7 by 8:10.

 

[kuruman]

No, that's a different question.
At 8:05 there may already be 3, 4, 5, ... people have visited the shop. We need the probabilities of those, and for each the probability that we haven't reached 7 by 8:10.

Yes it is a different question. I think the right question is "Find the probability that exactly 3 people walk in from 8:00 to 8:05 and no more than 3 walk in from 8:05 to 8:10."

 

[haruspex]

Yes it is a different question. I think the right question is "Find the probability that exactly 3 people walk in from 8:00 to 8:05 and no more than 3 walk in from 8:05 to 8:10."

I do not see that it can be that. We are told at least three have come in, so it is a conditional probability, as stated in the thread title, not a joint probability.

 

[kuruman]

I do not see that it can be that. We are told at least three have come in, so it is a conditional probability, as stated in the thread title, not a joint probability.

In retrospect, I don't see that it can be that either. "At least three" $\neq$ "exactly three". All I can say to my defense is that I parsed it incorrectly the first time I read it and then kept repeating the same error. Thanks.

 

[scottdave]

I read it as "equal to 3" in the first 5 minutes, as well. And was thinking that for a while.