Repeated trials

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zakhaev

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A shipment consists of 100 units. and the probability that a unit is defective is 0.2. We select 12 units; find the probability p that 3 of them are defective.
 
zakhaev said:
A shipment consists of 100 units. and the probability that a unit is defective is 0.2. We select 12 units; find the probability p that 3 of them are defective.
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Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this problem
 
You forgot to define YOUR variables! What are P1, D and A?

You multiply your work by 12/100. I assume that 12 is the number of items you pick and 100 is the number of items you pick the 12 from. Now if 100 was really 10,000 then according to your work, the answer would be smaller than what you got as multiplying a (positive) number by 12/10,000 yields a smaller number than multiplying this same number by 12/100. Does that make sense? Suppose the company make 1million items per day. Suppose before boxing the items they pick 12 vs boxing 100 items in each box, picking one box and choosing 12 from this box--should the probability of getting 3 defective be the same or not depending on which way you pick the 12 items to be tested?????
 
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Jomo said:
You forgot to define YOUR variables! What are P1, D and A?

You multiply your work by 12/100. I assume that 12 is the number of items you pick and 100 is the number of items you pick the 12 from. Now if 100 was really 10,000 then according to your work, the answer would be smaller than what you got as multiplying a (positive) number by 12/10,000 yields a smaller number than multiplying this same number by 12/100. Does that make sense? Suppose the company make 1million items per day. Suppose before boxing the items they pick 12 vs boxing 100 items in each box, picking one box and choosing 12 from this box--should the probability of getting 3 defective be the same or not depending on which way you pick the 12 items to be tested?????
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Thanks for your comment. So you are saying, it is not important how many items we have in general and the binomial distribution is enough?
 
zakhaev said:
A shipment consists of 100 units. and the probability that a unit is defective is 0.2. We select 12 units; find the probability p that 3 of them are defective.
Click to expand...
The title of your thread is Repeated Trials. In what sense do you mean repeated?
Do we know that there are twenty defective units in each shipment of one hundred?
If we select twelve at random, is the question about exactly three being defective or at least(at most) three?
 
[MATH]n = 12[/MATH] is a sample of the [MATH]100[/MATH] units

take the sample, ignore the [MATH]100[/MATH], and work normally with binomial distribution

[MATH]n = 12[/MATH][MATH]x = 3[/MATH][MATH]p(X=3) = \ ?[/MATH]
 
You say that the probability any one item is defective is 0.2. That's a bit odd. I would be inclined to think that means that 20 of the 100 items are defective and 80 are not. So the probability the first one is defective is 0.2. But then there are only 99 items left, 19 of them defective and the probability the next item is defective is 19/99, NOT 0.2!

IF there were an ifinite number of items to begin with (or, practically, a large number compared to the 12 samples) then we could use the binomial distribution. I am not sure that 100 is "large compared to 12"! But if we can use 0.2 for all choices then the probability the first three are defective is 0.2(0.2)(0.2)= 0.2^3. The probability the next 9 are NOT defective is 0.8(0.8)(0.8)(0.8)(0.8)(0.8)(0.8)(0.8)(0.8)= 0.8^9. The probability of "3 defective and 9 are not, in that order", is (0.2)^3(0.8)^9. There are \(\displaystyle \begin{pmatrix}12 \\ 3\end{pmatrix}= \frac{12!}{3! 9!}\) different orders so the binomial distribution gives \(\displaystyle \frac{12!}{3! 9!}(0.2)^3(0.2)^9\).​
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A more accurate way to do this is to say that the probability the first three items are defective and the last nine are not is \(\displaystyle \frac{20}{100}\frac{19}{199}\frac{18}{198}\frac{17}{197}\frac{80}{196}\frac{79}{195}\cdot\cdot\cdot\frac{2}{4}\frac{1}{3}\) and that must be multiplied by \(\displaystyle \frac{12!}{3!9!}\) to account for the different orders.​
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For a reasonably large total number of items, the soution from the binomial distribution will approximate that.​
 
Halls of Ivy is obviously correct IF THE ITEMS ARE CHOSEN WITHOUT REPLACEMENT. Selecting without replacement means that the trials are not independent, and independence is one of the requirements for using the binomial distribution.

If, however, the items are chosen with replacement, then the answer is indeed the binomial distribution because the trials are independent:

[MATH]\dbinom{12}{3} * (0.2)^3 * (0.8)^9 \approx 23.6\%.[/MATH]
Of course, (as pka pointed out, these answers are correct only if the question is asking for the probability of exactly 3 defectives. If the question is asking for the probability of at least three defectives, then we have an entirely different problem with different answers.

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zakhaev said:
Thanks for your comment. So you are saying, it is not important how many items we have in general and the binomial distribution is enough?
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No, you missed my point. It is not what I think. It is what you think!

BTW, you still forgot to define your variables. If you think that it is not a big deal, then you are certainly wrong. If I was having an off day when grading and saw your answer I'd give you 0 points for not defining your variables. If I was having a great day, then I would try to understand what your variables mean (and I am a terrible mind reader)and if I couldn't then you would get 0 points. Why not give the greater a chance to understand your work especially since it may be wrong. When you do the work wrong, then it hard to try to figure out what you meant to say when you did not define your variables
 
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