Probability exercise - stuck

[paveer]

I don't have any clue how to solve this exercise:

- The supplier knows that there is a 0.3 chance that customers will buy a certain product.
- If the supplier only has 3 pieces of that product in stock, what is the probability that less than five customers will be needed to use all the stock.

tx in advance for any suggestion

 

[HallsofIvy]

Are we to assume that a given customer will not buy more than one piece of that stock? If so then this is a binomial distribution with probability 0.3 of "success" and 0.7 of "failure". The probability that three of the first five customers will buy the stock and the other two won't is $\displaystyle \begin{pmatrix}5 \\ 3 \end{pmatrix}(0.3^3)(0.7^2)= (10)(0.027)(0.49)= 0.1323$.

 

[pka]

- The supplier knows that there is a 0.3 chance that customers will buy a certain product.
- If the supplier only has 3 pieces of that product in stock, what is the probability that less than five customers will be needed to use all the stock.

I wonder if Halls has done what is intended. But it is not what is written.
As I read it the question is: what is the probability that the stock of three will be exhausted by the third or fourth customer?

 

[JayJay]

I agree the problem, as written, is asking about stock remaining after four customers have been dealt with.

I see some ambiguity with "there is a 0.3 chance that customers will buy a certain product"; better would be "there is a 0.3 chance that a customer will want to buy a certain product". I think Hall's solution overlooks the case where the last customer(s) want to buy the item but cannot because all three originally in stock have been sold.

Probability that three of the four customers want to buy the item 4C3 x .3^3 x .7 = 0.0756
Probability that four of the four customers want to buy the item 4C4 x .3^4 = 0.0081
Probability that 3 items have been sold after four customers = 0.0756 + 0.0081 = 0.0837

 

[Steven G]

You need to be careful with this problem
You can sell the three items with just three customers.
You can sell the three items with four customers. Here is where you need to be careful. Selling the three items with four customers includes the cases where you sell the three items with just three customers. To avoid double counting you need for that third item to be purchased from the fourth customer.


Can you try to find the three probabilities (selling all 3 to 3 customers and selling all 3 to 4 customers where the 3rd is sold to the 4th customer

 

[Steven G]

Are we to assume that a given customer will not buy more than one piece of that stock? If so then this is a binomial distribution with probability 0.3 of "success" and 0.7 of "failure". The probability that three of the first five customers will buy the stock and the other two won't is $\displaystyle \begin{pmatrix}5 \\ 3 \end{pmatrix}(0.3^3)(0.7^2)= (10)(0.027)(0.49)= 0.1323$.

Prof Halls I have a problem with what you are doing. Can you please tell me if I am wrong? If the 3 items are purchased by the first 3 customers then wouldn't the probability of the last two persons purchasing the item be zero. I suspect that I am correct there. My concern is that this fact will make your calculation wrong(??)

 

[pka]

I think Hall's solution overlooks the case where the last customer(s) want to buy the item but cannot because all three originally in stock have been sold.
Probability that three of the four customers want to buy the item 4C3 x .3^3 x .7 = 0.0756
Probability that four of the four customers want to buy the item 4C4 x .3^4 = 0.0081
Probability that 3 items have been sold after four customers = 0.0756 + 0.0081 = 0.0837

First, did you retype the question form the original? Did you translate from another language?
If yes to any of those the try to post as near to an original as possible.
Now why do you add the $0.0081~?$ It says "the probability that less than five customers will be needed to use all the stock."
I don't read that as wanting to buy but none availed. The wording says that the third or fourth customer buys the last of the supply?
@Jomo the question says "what is the probability that less than five customers will be needed to use all the stock. "

 

[JayJay]

It is difficult to know to whom the above responses are directed. Before I comment further, can we establish whether we are thinking of four customers or five.

 

[pka]

It is difficult to know to whom the above responses are directed. Before I comment further, can we establish whether we are thinking of four customers or five.

I requested that the original poster answer. Her/his original post says "less than five customers will be needed to use all the stock".
Good form would have be to say "at most four" or "no more than four". But as I said in reply #3, I suspect the question in fact means five customers.

 

[paveer]

The question was translated to english, my apologies if i was not clear enough.
I assume in the original text that each customer would order only one type of product.
so the question was: what is the probability that less than five customers, for example 3 or 4, would consume the complete inventory.
Answer in our book is 0,0837 which matches with the explanation provided by Jay.
tx for all your valuable inputs!

 

[JayJay]

Paveer - happy to help!

Now why do you add the $0.0081~?$

To include the case where all four customers want to buy a piece.

Longer analysis -

Prob none buy a piece - 4C0 x .7^4 x .3^0 = 0.2401
Prob one buys a piece - 4C1 x .7^3 x .3^1 = 0.4116
Prob two each buy a piece - 4C2 x .7^2 x .3^2 = 0.2646
P1, prob at least one piece left after four customers -
0.2401 + 0.4116 + 0.2646 = 0.9163

P2, prob (no pieces left) - 4C3 x .7^1 x .3^3 = 0.0756 (as per HallsofIvy adapted fpr 4 customers)

P1 + P2 = 0.9163 + 0.0756 = 0.9919
But P1 + P2 should add to 1.0000
How do you explain the difference 0.0081?

 

[Steven G]

Here is how I would solve this.

p(1st three customer buy all three items) = .3^3
p(1st three customers bought 2 items and the 4th customer did buy the last one). So 2 of the 1st 3 customers bought the item and the 4th customer bought the item. This probability is ₃C₂*(.3)^3(.7)
Just add these two values as the intersection is the empty set and you'll have your answer.

 

[pka]

Here is how I would solve this.
p(1st three customer buy all three items) = .3^3
p(1st three customers bought 2 items and the 4th customer did buy the last one). So 2 of the 1st 3 customers bought the item and the 4th customer bought the item. This probability is ₃C₂*(.3)^3(.7)
Just add these two values as the intersection is the empty set and you'll have your answer.

I agree with the above solution. I think that it is consistent with the English translation of the problem.
Which asks about the probability that the supply of three items in stock is depleted by the first four customers.
We do assume that it is at most one per customer. Moreover, there is nothing stated nor implied about costumers' desire to buy even after the stock is depleted.