Probability: 94% of bolts in tray within diameter tolerance

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roburtr

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Hello guys,

I am struggling with a probability problem for my assignment, the problem text is:

"You visit your Manufacturing division, which has a machine producing metal bolts. In a tray of these bolts, 94% are within the allowable diameter tolerance value. The remainder exceed the tolerance. You withdraw six bolts at random from the tray. Determine the probabilities that;

(i) Two of the six bolts exceed the diameter.
(ii) More than two of the six bolts exceed the diameter."

On the firs request (i) I have come up with a solution which I am 98% sure is wrong.I made a table with all the possible combinations. Let's say: A=Bolts with exceeded diameter=20/100
and
B=Bolts with good diameter=80/100

So...

1) A A B B B B______________20/100*20/99*80/98*80/97*80/96*80/95=0.019
2) A B A B B B______________20/100*80/99*20/98*80/97*80/96*80/95=0.019
3) A B B A B B______________20/100*80/99*80/98*20/97*80/96*80/95=0.019
..........................................................................................................................
..........................................................................................................................
15) B B B A A B______________80/100*80/99*80/98*20/97*20/96*80/95=0.019

So,

0.019*15=0.28, So The probability that two of six bolts exceed the diameter is 0.28.

Is that correct?

Now if this is correct then at second request (ii) I am in trouble, because the table is huge and there are a lot of combinations. Now talking about combinations I think I need to use combinations formulae to solve the problem. Am I right? I apologise if What my answer to the problem doesn't make sense, but I am struggling with math. :confused::confused::confused::confused::confused::confused:.

Thank you very much.
 
Hi robutr

Where the heck do A = 20/100 and B = 80/100 come from? The probability of success of 94%, not 80%

You are right that a combination formula does come into play, because you need to compute the number of distinct ways of choosing two bolts out of the six to be bad.

The general solution to this type of problem is given by the binomial probability distribution. I just wrote an explanation of the binomial distribution in response to another thread. Check it out:

https://www.freemathhelp.com/forum/threads/111037-Calculating-chance-of-life-on-another-planet

That guy was trying to figure out the probability that a certain number of planets would have life, out of a sample. But it's the same problem. In your case you have N=6\displaystyle N =6 trials with a success rate of p=0.94\displaystyle p = 0.94 and a number of successes of k=4\displaystyle k = 4.
 
Hello , was wondering if you could please be of some assistance ( 40 year old offshore construction supervisor) I am currently working on the same problem and having difficultly calculating the final formula. as below is what i have so far

your Manufacturing division, which has a machine producing metal bolts. In a tray of these bolts, 94% are within the allowable diameter tolerance value. The remainder exceed the tolerance. You withdraw six bolts at random from the tray. Determine the probabilities that;

q=.94
p=.06
?(2) = ?(? − 1) ??−2?2
2
?(2) = 6(6 − 1) .94(6-2).06(2)
2
P(2)=(15)(.734)(.0036)

(i) Two of the six bolts exceed the diameter. = P2=.0396


?(>2)=1−{?(0)+?(1)+?(2)}
?(> 2) = 1 − {0.946 + (6)(0.945 )(0.06) + 0.0396} Really struggling to calculate this formula out , been a very long time since i have used a scientific calculator. Any help you can be with this would be greatly appreciated.

(ii) More than two of the six bolts exceed the diameter." ????
 
These are the answers i got i havent got a clue if they're correct. What do you think?
 

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Probability of success on a single trial = p

Probability of failure on a single trial = (1 - p).

Trials are independent.

probability of m successes in n trials [MATH]= \dbinom{n}{m} * p^m * (1 - p)^{(n-m)}.[/MATH]
A formula that comes up time and time again in probability.
 
S93 said:
So the formula i've used has no relevance what so ever?
Click to expand...
It appears that you are trying to do something with the Poisson distribution. I have not worked with the Poisson in ages. But this is just a straight binomial distribution problem. Have you studied that?

By the way, do you really believe that the probability of getting a defective bolt in one trial is 0.96 if the frequency of good bolts is 94%?
 
Yes that's what i found in my notes from what qe learned in class.

No i think it must be wrong but going with he information I've got and how I've been taught that's the answer I've come up with. If I'm honest it makes no sense what so ever to me
 
The binomial distribution applies whenever the probability of "success" for a single RANDOM trial is known, and it is known that successive trials are independent. It is traditional to label this probability p, and the probability of failure on a single trial as q = 1 - p. The important point is that p is the same for each individual trial.

To do this kind of problem, you need to know whether successive trials are independent and so require information, explicit or implicit, on how the experiment is being run. The roulette wheel is not affected by what it did on the previous spin. What cards remain in the dealer's shoe are affected by what cards have already been played. The information given in the initial post is insufficient to figure that out. Post # 2 guessed that they are independent. In post 5, I gave you the formula to use if the guess in post 2 is correct. It is very straight forward. If the trials are not independent, the computations get a bit more complex.

What is "success" for this problem: getting a defective bolt.

You know that 94 bolts in the tray are not defective and that there are 100 bolts in the tray. So when you pick the first bolt, what is the probability that it is defective?

At this point, we need to ask how this test is being conducted.

Maybe the problem says that you pick a bolt at random, note whether it is defective, throw the bolt back in the tray, shuffle them all around and pick at random again. Are successive trials independent in that case?

Maybe the problem says that you pick a bolt at random, note whether or not it is defective, and put it to the side. Then you pick another bolt, and so on. Are successive trials independent in that case?
 
JeffM said:
The binomial distribution applies whenever the probability of "success" for a single RANDOM trial is known, and it is known that successive trials are independent. It is traditional to label this probability p, and the probability of failure on a single trial as q = 1 - p. The important point is that p is the same for each individual trial.

To do this kind of problem, you need to know whether successive trials are independent and so require information, explicit or implicit, on how the experiment is being run. The roulette wheel is not affected by what it did on the previous spin. What cards remain in the dealer's shoe are affected by what cards have already been played. The information given in the initial post is insufficient to figure that out. Post # 2 guessed that they are independent. In post 5, I gave you the formula to use if the guess in post 2 is correct. It is very straight forward. If the trials are not independent, the computations get a bit more complex.

What is "success" for this problem: getting a defective bolt.

You know that 94 bolts in the tray are not defective and that there are 100 bolts in the tray. So when you pick the first bolt, what is the probability that it is defective?

At this point, we need to ask how this test is being conducted.

Maybe the problem says that you pick a bolt at random, note whether it is defective, throw the bolt back in the tray, shuffle them all around and pick at random again. Are successive trials independent in that case?

Maybe the problem says that you pick a bolt at random, note whether or not it is defective, and put it to the side. Then you pick another bolt, and so on. Are successive trials independent in that case?
Click to expand...
why are you sayingthere are 94 bolts in the tray. it is given as a percentage. the amount is unknown
 
The unknown amount is utterly irrelevant. We can turn percents into relative amounts simply by multiplying by 100 and make the problem concrete.
 
Sparky1977 said:
Thank for your response Jeff. It's been confusing me. How do I solve it?
Click to expand...
Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

​
Please share your work/thoughts about this problem.
 
As I said in an earlier post, the information given by the original poster is insufficient to determine the answer. It depends on how the test is being conducted. In addition to liking to know what you have tried or at least thought about, we also really need the complete exact wording of the problem.

Also take a look at post # 9.
 
At the moment i'm struggling with what equation to use and thus have no working. I was considering a binomial distribution and was learning that. However, one of the conditions of BD is that the trials are independant. In this problem i dont think they are.

The exact wording of the problem is in post #1.
Also, when you suggest turning a percentage into a quantity to make it concrete, correct me please if i'm wrong but wouldn't that then suggest that 1 bolt would be 1% and therefore skew results if it turned out this wasn't the case.
Many thanks for your time and patience,
Martin.
 
If that is the exact and complete wording of the problem, it is poorly written. It is impossible to tell from that wording whether the trials are independent.

You are correct that the binomial distribution is not relevant if the trials are not independent.

Furthermore, you are correct that the size of the universe DOES affect the probability if the trials. I was wrong. But given that, the problem is even written more vaguely because it does not give the size of the universe.

When I say that exactly 94% of the bolts are within tolerance, that must mean that the number of bolts in the tray is some multiple of 50 bolts, but we do not know which multiple as the problem seems to be worded. (It has to be a multiple of 50 to avoid talking about fractional bolts.) Furthermore, the problem does not make sense if there are only 50 bolts in the tray because there is no possibility that all six of our sample can be defective: there are only three defective in the tray because 3 = 0.06 * 50. So it is plausible to assume that whoever wrote the problem meant that the tray contains 100 bolts. Let's do the problem on that basis initially. .

There are two ways to solve this kind of problem. Let's do an easier problem. What is the probability that exactly one bolt chosen is defective?

Type A: That the first bolt chosen is defective and all succeeding ones are not defective has a probability of

[MATH]\dfrac{6}{100} * \dfrac{94}{99} * \dfrac{93}{98} * \dfrac{92}{97} * \dfrac{91}{96} * \dfrac{90}{95}.[/MATH]
We used the definition of conditional probability. Do you see how?

That the first bolt chosen is not defective, the second bolt is defective, and all succeeding ones are not defective has a probability of

[MATH]\dfrac{94}{100} * \dfrac{6}{99} * \dfrac{93}{98} * \dfrac{92}{97} * \dfrac{91}{96} * \dfrac{90}{95}.[/MATH]
We do that for each bolt and add up the probabilities, which are mutually exclusive.

[MATH]\dfrac{6}{100} * \dfrac{94}{99} * \dfrac{93}{98} * \dfrac{92}{97} * \dfrac{91}{96} * \dfrac{90}{95} +\\ \dfrac{94}{100} * \dfrac{6}{99} * \dfrac{93}{98} * \dfrac{92}{97} * \dfrac{91}{96} * \dfrac{90}{95} + \\ \dfrac{94}{100} * \dfrac{93}{99} * \dfrac{6}{98} * \dfrac{92}{97} * \dfrac{91}{96} * \dfrac{90}{95} + \\ \dfrac{94}{100} * \dfrac{93}{99} * \dfrac{92}{98} * \dfrac{6}{97} * \dfrac{91}{96} * \dfrac{90}{95} + \\ \dfrac{94}{100} * \dfrac{93}{99} * \dfrac{92}{98} * \dfrac{91}{97} * \dfrac{6}{96} * \dfrac{90}{95} + \\ \dfrac{94}{100} * \dfrac{93}{99} * \dfrac{23}{98} * \dfrac{91}{97} * \dfrac{90}{96} * \dfrac{6}{95} = \\ \dfrac{6 * (6 * 94 * 93 * 92 * 91 * 90}{100 * 99 * 98 * 97 * 96 * 95} \approx 27.63\%[/MATH]
Type B: we realize that we do not care about the order in which the bolts are drawn and use what we know about combinations.

How many ways can we draw exactly 1 from 6. How many ways can we draw exactly 5 from 94. How many ways can we draw 6 from 100.

[MATH]\dfrac{\dbinom{6}{1} * \dbinom{94}{5}}{\dbinom{100}{6}} = \dfrac{\dfrac{6!}{1! * 5!} * \dfrac{94!}{5! * 89!}}{\dfrac{100!}{6! * 94!}} =[/MATH]
[MATH]\dfrac{6!}{1! * 5!} * \dfrac{94!}{5! * 89!} * \dfrac{6! * 94!}{100!} =[/MATH]
[MATH]6 * \dfrac{94 * 93 * 92 * 91 * 90}{5 * 4 * 3 * 2} * \dfrac{6 * 5 * 4 * 3 * 2}{100 * 99 * 98 * 97 * 96 * 95} =[/MATH]
[MATH]\dfrac{6 * (6 * 94 * 93 * 92 * 91 * 90)}{100 * 99 * 98 * 97 * 96 * 95} \approx 27.63\%.[/MATH]
Type B is a little harder to grasp conceptually, but it is less prone to error and quicker to compute.

Using the assumption that there are 100 bolts on the tray, give the actual problem a go.
 
Thank you sooo much for your time and kindness. I now have a formula for it which has been verified. I shall share it with you tomorrow.
Martin
 
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