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Probability Question (theoretical aid)

  • Thread starter Centurion1
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  • #1
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Homework Statement


2. Photocards-R’-Us is a photo card printer. For this year, they have found that 10% of the cards they make are of inferior quality (blurred picture, faded colors, etc.) To attract customers to have their holiday cards printed with them, they are offering a 50% discount to customers who order a box of 50 cards if 10 of the cards in the box are of inferior quality. Manny is going to order a box of 50 Christmas photo cards at Photocards-R’-Us. What is the probability that he will receive a 50% discount on his holiday cards?

Homework Equations



none given

The Attempt at a Solution



I believe it to be something like a binomial expansion or a bayes thereom. Once again i am unsure of how to start the problem rather than the actual math involved. I am just not sure of what to employ.
 

Answers and Replies

  • #2
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n/a.

Actually I tried doing a bernoulli process which seems logical and I ended up with a result of 1.5%. I was wondering if this was accurate. I would write out the process but i dont really have the needed icons. It is like most of stats mostly basic math though with some factorials i leave to the calculator ;)
 
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  • #3
Ray Vickson
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n/a.

Actually I tried doing a bernoulli process which seems logical and I ended up with a result of 1.5%. I was wondering if this was accurate. I would write out the process but i dont really have the needed icons. It is like most of stats mostly basic math though with some factorials i leave to the calculator ;)
You should show your work, because I get the different answer Pr{discount} = 2.45%, using standard formulas for the appropriate distribution, and using Maple to do the computations. (I have assumed the discount occurs if the box contains 10 OR MORE inferior cards, rather than 10 exactly.)

RGV
 
  • #4
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hmmmm..... ok i will try.

Bernoulli

P(x)= n!/x!(n-x)! * τ^x(1-τ)^n-x

n= 50
τ= .10
x= 10

plug and chug.
 
  • #5
Ray Vickson
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hmmmm..... ok i will try.

Bernoulli

P(x)= n!/x!(n-x)! * τ^x(1-τ)^n-x

n= 50
τ= .10
x= 10

plug and chug.
1) That is a Binomial distribution, not a Bernoulli.
2) Plug what, and chug how? Do you have just one value of x, or several, and what are they?

RGV
 
  • #6
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it calls it bernoulli.

all i have to do is plug in the values.
 
  • #7
Ray Vickson
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it calls it bernoulli.

all i have to do is plug in the values.
A Bernoulli random variable takes only 2 values: 0 and 1. A Bernoulli process is a sequence of independent Bernoulli random variables. A binomial random variable is the number of "1's" in a Bernoulli process. The only time a Bernoulli random variable equals a Binomial random variable is when n = 1. I don't know what the "it" is that you cite, but if it calls a binomial a bernoulli then it is just plain wrong.

Anyway, you have still not answered my question: what value or values of x should you use? Considering that you got the wrong answer, I would have thought you would have some interest in knowing where your errors lie.

RGV
 
  • #8
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Values of x in what formula? The one I used?

I called it a bernoulli becasue in my statistics book and because in class we used that formula and called it bernoulli
 
  • #9
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bump..... trying to understand im not trying to be difficult.
 
  • #10
Ray Vickson
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bump..... trying to understand im not trying to be difficult.
OK, let me ask once again. You have a formula P = [n!/x!(n-x)!] T^x (1-T^(n-x), and you have said that n = 50 and T = 0.1. OK so far? But, you want to compute a probability numerically, so you need some numerical value of x. All I have been asking you to tell me is what the numerical value of x is in this problem. Let me remind you: YOU were the one who wrote the formula originally, so I guess you had some x in mind when you wrote it. Also, you said you had computed a numerical answer (wrongly!), so somehow you used some numbers when doing that. What numbers did you use?

RGV
 
  • #11
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n= 50
τ= .10
x= 10

those are the numbers i plugged in.

x= number of successes. int his case a success is if the card is defective.
 
  • #12
Ray Vickson
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n= 50
τ= .10
x= 10

those are the numbers i plugged in.

x= number of successes. int his case a success is if the card is defective.
So, your calculation is the probability of having *exactly* 10 defectives in a batch of 50. You must be claiming that if there are 11 or 12 or 13 or ... or 50 defectives, the customer does not get a discount. He/she only gets it for exactly 10 defectives.

RGV
 
  • #13
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according to the question in the way it is worded it only is 10.
 
  • #14
Ray Vickson
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according to the question in the way it is worded it only is 10.
I agree that is the wording, but I would bet that is not the meaning. If it were me, I would do it both ways and hand in two clearly-labelled answers.

RGV
 
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  • #15
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If 11 cards are of inferior quality, that implies that 10 cards are of inferior quality as well. No need to do it twice.
 
  • #16
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i asked my professor. She said that it was the 1.5% answer. It may be that she is filipino some of the other questions are somewhat vague in the wording as well.
 
  • #17
Ray Vickson
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If 11 cards are of inferior quality, that implies that 10 cards are of inferior quality as well. No need to do it twice.
This is incorrect. If *exactly* 11 cards are of inferior quantity then it is not true to say that exactly 10 are of inferior quality. The formula p(x) = C(50,x)(.1)^x (.9)^(50-x) gives, for x= 11, the prob. of exactly 11 defectives.

RGV
 
  • #18
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If 11 cards are of inferior quality, that implies that 10 cards are of inferior quality as well. No need to do it twice.
This is incorrect. If *exactly* 11 cards are of inferior quantity then it is not true to say that exactly 10 are of inferior quality. The formula p(x) = C(50,x)(.1)^x (.9)^(50-x) gives, for x= 11, the prob. of exactly 11 defectives.
they are offering a 50% discount to customers who order a box of 50 cards if 10 of the cards in the box are of inferior quality.


Well if you pull the word exactly out of thin air and insert it wherever you want into the question, then yes I'm wrong.
 
  • #19
Ray Vickson
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I am not pulling words out of the air. I am saying exactly what a certain FORMULA is computing, and am saying exactly how terms are used in *probability theory*, not in everyday conversation.

RGV
 

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