Another statistics question(from todays midterm)

[pyrate83]

See if you can solve it. I had trouble solving it but I wanted some input. Might help me on upcoming problems and tests.

A company that produces widgets knows that 2% of the widgets produced are defective. If 4 widgets are chosen randomly, what is the probability that at least 1 of those 4 will be defective?

[shelbyGT]

You need to know how many widgets they produce a day to be able to tell the chances that one of your 4 random sample is going to be defective.

And don't make me retake statistics class.... enough Ho and Ha and null hypothesis and crap... argh, I can't stand it!

[pyrate83]

You can make up an arbitrary number. This wasn't multiple choice or anything, it was work out the problem and give an answer.

[shelbyGT]

All I know is that the percentage that at least one of them is defective is different than if you asked for the percentage that only 1 of them is defective.

[pyrate83]

Yeah. Damn thing was worth 15 points is all I can say.

[Cobra]

The probability is 8%.

If you randomly select ONE widget, you have been told that it stands a 2% chance of being defective. Therefore, if you choose TWO then the chances of one of them being defective double.

Therefore:

Chance of one or more units being defective = probablility of one being defective x number of units in sample

[pyrate83]

Damn 8% is what I almost wrote down.

I took 1/4x1/4x1/4x1/4 and used that number as the probability instead.

Oh well, now I know...but damn. At least partial credit is given so maybe I'll get something for my effort.

[shelbyGT]

Are you sure about that?

I don't know, it's been a little bit since I was in college, but that just doesn't seem right in my head.

[pyrate83]

Are you sure about that?

I don't know, it's been a little bit since I was in college, but that just doesn't seem right in my head.

I'm not sure about anything in that class.

[shelbyGT]

I don't know your answer, but I just don't think it's 8%, that's all. The chances that 2 out of 4 are defective is not double that the chances that 1 out of 4 are. If anything, the chances of 2 out of 4 are half of 1 of 4.

At least that's how my brain is operating right now. Too much halogen light for today, it fries the brain.

[ViRaL1]

I think I'd go with 8%. There's a 2 percent chance for any 1 to be bad. The more you add the greater the chances of any one of them being bad. If we were talking about the chances of all of them being bad, then we'd divide instead of multiply.

[shelbyGT]

By the way you all are thinking, wouldn't you just stick with 2%? If the average is 2%, the more you add... the average doesn't change. I still don't think 2% is right, but... that's the train of thought you all are taking, so why wouldn't you just stick with 2%?

[shelbyGT]

If only I could bring in my old operations management teacher in here, he'd tell you all about it!

[shelbyGT]

taking a random sample of 4 and saying that the chance that at least two of them are defective is higher than the chance of 1 of them being defective... just doesn't seem right! I want the answer now, damnit!

[ViRaL1]

No, we're still looking for just ONE defective unit, we just have more to choose from.

[shelbyGT]

the question was stated that "at least one of the 4", so technically if it is a representative sample, it would be 2% that just one is defective of the 4. If you are trying to find the percentage that 2 out of those 4 are defective, that's LESS likely.

[ViRaL1]

Right, that's what I was saying way up there about dividing if were looking for more than 1 bad one.

[pyrate83]

Now I'm really lost.

[shelbyGT]

So if you are agreeing that the chances that 2 being defective are less than one... why did you agree with the 8%?

[pyrate83]

I didn't necessarily agree that 8% was right. Again, I didn't put that answer down on the test.

[ViRaL1]

Think of it like this. 2% of all widgets are defective. Once you have 100 widgets, at least from a statistical standpoint, (although it's still probability and not quite reality) there's a 100% chance that two of them are defective. As the number of widgets increases, the chances of at least one of them being defective increases.

[shelbyGT]

Hmm, I can see that point, but that's not the direction I thought the question was going. Aww well.

It's a very badly worded question. If they had stated the production capacity of widgets in a day, we could have given a real answer.

[pyrate83]

Yes I do understand that. That would make sense.

[Cobra]

shelbyGT - they are asking what the probability of one or more widgets being bad is, not what the chances that just one is. There is a subtle wording difference, but a big mathematical difference.

[ViRaL1]

Well, I THINK I'm on the right track but I may need to rethink my numbers.

[shelbyGT]

I see your point, and I was just going in a different direction. They are asking the chances that 1 out of a random 4 sample from X number produced is going to be defective. Yes, the larger the sample is, the larger the probability that at least one is defective. So, technically we were both right. However, to have given any kind of answer, they needed to state how many widgets can be produced in a run with that margin of error.

[ViRaL1]

Now I'm thinking more along the lines of 12.5%...
2/100 = 1/50, 4 widgets = 4/50 = 12.5. I never took stats though, so I could be way off. Stats are kind of funny, like the idea of a 100% chance of 2 being defective. The reality is that coin tosses don't go HEADS, TAILS, HEADS, TAILS, TAILS, HEADS. Supposedly they'll converge to 50% somewhere way down the road.

[pyrate83]

Now I'm thinking more along the lines of 12.5%...
2/100 = 1/50, 4 widgets = 4/50 = 12.5. I never took stats though, so I could be way off. Stats are kind of funny, like the idea of a 100% chance of 2 being defective. The reality is that coin tosses don't go HEADS, TAILS, HEADS, TAILS, TAILS, HEADS. Supposedly they'll converge to 50% somewhere way down the road.

I just want to get through with a C and I'll be happy. This shit just gives me a serious headache.

[Cobra]

Yeah... I've messed up...

Sorry!

[jimmalenko]

2% is 1 in every 50, or 0.02. But with every "normal" one that is drawn, the probability that a defective one will be next increases. From what I can remember, you gotta use factorials and shit to work this out ...

http://www.saliu.com/theory-of-probability.html

BDF = (N!/((N — M)! * M!)) * pM * (1 — p)^(N — M)

BDF = the binomial distribution probability;
p = the individual probability of the phenomenon (e.g. p = 0.5 to get tails in coin tossing);
M = the exact number of successes (e.g. exactly 5 tails in 10 coin tosses);
N = the number of trials (e.g. exactly 5 tails in 10 coin tosses = number of trials).

p=0.02
M=1
N=4

You can work out the rest ...

.. then because the question asks for one or more successes, you gotta work it out for M=1,2,3 & 4 then add them all together to get the probability that at least one will be defective ...