Don`t shop in this kind of supermarket! Prob. random customer has error in bill

[fibster]

There are 5 working cash desks in supermarket. It`s been observed that 21% of customers choose first desk, 18% - second, 20%- third, 22% -fourth, 19% - fifth.
First cashier makes mistake every 23. bill, second every 40. bill, third every 150. bill, fourth every 30. bill and fifth every 35. bill.
What is probabilty that random customer will have error in a bill?
How many percent of all mistaken bills are related to each cash desk?

 

[Dr.Peterson]

There are 5 working cash desks in supermarket. It`s been observed that 21% of customers choose first desk, 18% - second, 20%- third, 22% -fourth, 19% - fifth.
First cashier makes mistake every 23. bill, second every 40. bill, third every 150. bill, fourth every 30. bill and fifth every 35. bill.
What is probabilty that random customer will have error in a bill?
How many percent of all mistaken bills are related to each cash desk?

I don't shop there either, but that doesn't keep me from trying. (Much of the terminology sounds foreign to me.)

For each "desk" (A, B, C, D, E), we are given the probability that it is chosen, e.g. P(A), and the probability of an error, e.g. P(error | A).

Have you learned anything about compound probabilities? To answer the first question, you want P(error) = P(A)*P(error|A) + ...

For the second question, you want P(A | error), and so on.

Give it a try, and let us know where you get stuck. We need to see your work in order to know what help you need.

 

[pka]

There are 5 working cash desks in supermarket. It`s been observed that 21% of customers choose first desk, 18% - second, 20%- third, 22% -fourth, 19% - fifth.
First cashier makes mistake every 23. bill, second every 40. bill, third every 150. bill, fourth every 30. bill and fifth every 35. bill.
What is probabilty that random customer will have error in a bill?
How many percent of all mistaken bills are related to each cash desk?

Please explain exactly what "makes mistake every 23. bill" means.

 

[fibster]

Please explain exactly what "makes mistake every 23. bill" means.

It means that there is 1 error in 23 bills. Sorry about my poor English.

 

[Dr.Peterson]

Please explain exactly what "makes mistake every 23. bill" means.

My assumption is that it means something like "makes a mistake on every 23rd bill of sale (sales slip)". As I said, it sounds foreign; but I've seen the dot used for an ordinal in some countries.

 

[fibster]

I don't shop there either, but that doesn't keep me from trying. (Much of the terminology sounds foreign to me.)

For each "desk" (A, B, C, D, E), we are given the probability that it is chosen, e.g. P(A), and the probability of an error, e.g. P(error | A).

Have you learned anything about compound probabilities? To answer the first question, you want P(error) = P(A)*P(error|A) + ...

For the second question, you want P(A | error), and so on.

Give it a try, and let us know where you get stuck. We need to see your work in order to know what help you need.

1. ((21/100)*(1/23))+((18/100)*(1/40))+((20/100)*(1/150))+((22/100)*(1/30))+((19/100)*(1/35)) = 0.0277
2. (1/23) + (1/40) + (1/150) + (1/30) + (1/35) = 0,137 ~14%

Is that right?

 

[Dr.Peterson]

There are 5 working cash desks in supermarket. It`s been observed that 21% of customers choose first desk, 18% - second, 20%- third, 22% -fourth, 19% - fifth.
First cashier makes mistake every 23. bill, second every 40. bill, third every 150. bill, fourth every 30. bill and fifth every 35. bill.
What is probabilty that random customer will have error in a bill?
How many percent of all mistaken bills are related to each cash desk?

1. ((21/100)*(1/23))+((18/100)*(1/40))+((20/100)*(1/150))+((22/100)*(1/30))+((19/100)*(1/35)) = 0.0277
2. (1/23) + (1/40) + (1/150) + (1/30) + (1/35) = 0,137 ~14%

Is that right?

The first answer is good. The second is not.

Think about what I said. You want five different answers (one for each desk), the first one being P(A | error). How is that conditional probability defined? Isn't it P(A and error)/P(error)? In the first part, you found P(error) ...

 

[fibster]

The first answer is good. The second is not.

Think about what I said. You want five different answers (one for each desk), the first one being P(A | error). How is that conditional probability defined? Isn't it P(A and error)/P(error)? In the first part, you found P(error) ...

Thank you for helping me! I read the question in my native language and got the idea. Thank you for all of your support!

P(1)= ((21/100)*(1/23)) / (1/23))= 21%
P(2)=((18/100)*(1/40))/(1/40))= 18%
P(3)=((20/100)*(1/150))/(1/150))= 20%
P(4)=((22/100)*(1/30))/(1/30))= 22%
P(5)=((19/100)*(1/35))/(1/35)) = 19%

 

[Dr.Peterson]

Thank you for helping me! I read the question in my native language and got the idea. Thank you for all of your support!

P(1)= ((21/100)*(1/23)) / (1/23))= 21%
P(2)=((18/100)*(1/40))/(1/40))= 18%
P(3)=((20/100)*(1/150))/(1/150))= 20%
P(4)=((22/100)*(1/30))/(1/30))= 22%
P(5)=((19/100)*(1/35))/(1/35)) = 19%

Not quite. In effect, you just ignored the individual probabilities of an error for each desk, and gave back the percentages for choosing each desk.

P(error), in my terms, is the answer to the first question, not 1/23, etc., which are P(error | A), etc.

 

[JeffM]

Thank you for helping me! I read the question in my native language and got the idea. Thank you for all of your support!

P(1)= ((21/100)*(1/23)) / (1/23))= 21%
P(2)=((18/100)*(1/40))/(1/40))= 18%
P(3)=((20/100)*(1/150))/(1/150))= 20%
P(4)=((22/100)*(1/30))/(1/30))= 22%
P(5)=((19/100)*(1/35))/(1/35)) = 19%

Do you see that when you multiply by (1/23) and divide by (1/23), you are accomplishing nothing.

I am going to use \(\displaystyle D_i\) to mean that customer goes to the ith desk.

\(\displaystyle P(\text {Error } | \ D_1) = \dfrac{1}{23} \text { and } P(D_1) = \dfrac{1}{21}.\)

\(\displaystyle \therefore P(\text {Error } | \ D_1) * P(D_1) = \dfrac{1}{23} * \dfrac{1}{21} = \text { the probability of what?}.\)

\(\displaystyle \displaystyle \sum_{i=1}^5 P(\text {Error } | \ D_i) * P(D_i) = \text { the probability of what?}\)