Customers come into a store. $30\%$ of them make a purchase.
a) Calculate the probability that the second purchase is made by the sixth customer.
Answer: As the last purchase is made by the sixth customer,
we have a binomial with n=5 times the probability of the last purchase.
The probability is $C(5,1) (0.3) \cdot 0.7^4 \cdot 0.3$ = 0.108045
b)Calculate the expected value of the sequence number of the customer who makes the second purchase.
I have several questions related to part b):
1) What is a sequence number?
2) The question is part of the probability manual written by Abraham Weishauss. Weishauss answer is to find the expected value and add 2. My question is why? Weishauss does not explain that.
For the Negative Binomial Distribution, the expected value is $k\cdot(1-p)/p$.
For that reason the expected value is $2 \cdot (1-0.3)/0.3 = 14/3$
Adding 2 to that, Weishauss says that the answer is $20/3$
3) Weishauss explanation is "The negative binomial mean measures the number of customers between the second one and the one making the purchase. Therefore, we add 2 to the mean to get the sequence number of the customer making the purchase."
I do not understand this explanation. May you expand this answer?
I like to understand the ideas. I do not see the intuition behind this answer.
Best Answer
The indexed count in the sequence.
Where $p$ is the success rate, and $k$ is the count for successes, then $k(1-p)/p$ is the expected count of failures before success $\#k$. It does not include the successes.
So, adding the two successes gives us the count for trials until the second success.
I do not understand this explanation. May you expand this answer?
A Geometric distribution counting the failures before the first success has a mean of $(1-p)/p$.
So the expected count of failures before the first success, the first success, failures from then before the second success, and the second success is $2(1-p)/p+2$ or simply $2/p$.