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# Binomial and poisson quick question. watch

1. I've got the answer, but I am just wondering something here about how the answer should be acheived. But anyway, here's the question:

"A large number of screwdrivers from a trial production run is inspected. It is found that the cellulose acetate handles are defective on 1% and the chrome steel blades and defective on 1.5% of the screwdrivers, the defects occuring independently.
(i) what is the probability that a sample of 80 contains more than two defective screwdrivers?"

The answer I've got several ways, and the answer in the markscheme is 0.323. However, my several different answers aren't exactly the same but all round to 0.323 to 3 dp, so which of the following methods are correct?
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1)
The "brute force" way that I did which I am sure is correct is to work out the probability of none of the 80 being defective in either way + probability one out of 80 is defective handle but none out of 80 is defective with blades + etc etc then take one away from that answer. Is that way correct?

2)
Another way I thought is to realise that I want the probability of a screwdriver being defective and it doesn't matter how. So am I interested in P(AuB)? In which case, this is 0.01+0.015-0.01*0.015 = 0.02485. From here, do I simple use a binomial?
so like:
1- (0.97515^80+80*0.97515^79*0.0248 5+80c2*0.97515^78*0.02485^2)
but that = 0.3200200823. And I can't see why that isn't correct.
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3) if I do the binomial the same way but just get the probability of a screwdriver being defective by simply adding 0.01 and 0.015 it gets the answer in the markscheme :s. But that surely can't be correct because i've only added the probabilities together and havnt took into consideration the screwdrivers that have both defects.

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It's confusing me because I keep thinking "what if the % of different defects was really high like 90%" and then I get very confused because you don't know which 90% of each overlap and stuff, you know? By "overlap" I mean, screwdrivers that are defective becuase they have both defects. Anyway I'd appreciate some help, cheers.
2. I agree with your method (2). I don't think they worried about the two defects coinciding, but they should have done.
3. (Original post by DFranklin)
I agree with your method (2). I don't think they worried about the two defects coinciding, but they should have done.

so do you think it's the markscheme that's wrong?

After posting this I did my first method again and I realised that without rounding errors both methods get the same answer as well.
4. I think the markscheme is wrong. Did they expect you to use binomial or poisson, incidentally? In some senses, Poisson distributions model the scenario where "events are rare enough we don't worry about them happening twice", so I can see that putting a bias towards not considering coinciding defects. But (to me, at any rate) it's clear that (b) is actually the correct analysis.
5. (Original post by DFranklin)
I think the markscheme is wrong. Did they expect you to use binomial or poisson, incidentally? In some senses, Poisson distributions model the scenario where "events are rare enough we don't worry about them happening twice", so I can see that putting a bias towards not considering coinciding defects. But (to me, at any rate) it's clear that (b) is actually the correct analysis.

I think it's expecting poisson, but how would you go about doing it with that anyway? If you combine both possion distrubtions you're only looking at screwdrivers with both defects. How would you do it? the brute force way only with poisson instead?
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also, I was wondering if you could clear something up for me regarding events overlapping and stuff.
Consider, I don't know, 100 screwdrivers. 50% have defective handles and 50% have defective blades independent of each other.
So P(AnB) would be: 0.5*0.5 = 0.25, right?
So P(PuB) would be 0.5+0.5-0.25 = 0.75???
But isn't that saying that 75% are defective somehow and 25% are fine? How do you know? Isn't it possible that the 50% with defective handles are the exact same 50% with defective blades? Or perhaps every screwdriver is defective with either it having a defective blade or defective handle but no overlapping? I don't see how it makes sense.
6. The "natural" way to do it with Poisson would be to say

#of defect A is Pois(80 * 0.01) = Pois(0.8)
#of defect B is Pois(80 * 0.015) = Pois(1.2)

To get the distribution of the sum of two Poissons, you just add the parameters. That is:

(# of defect A) + (# of defect B) is Pois(0.8+1.2) = Pois(2).

The problem is (as we have discussed), the total number of defects is NOT going to be quite the same as the total number of defective screwdrivers. Some screwdrivers will have two defects.
7. (Original post by Tallon)
Consider, I don't know, 100 screwdrivers. 50% have defective handles and 50% have defective blades independent of each other.
So P(AnB) would be: 0.5*0.5 = 0.25, right?
So P(PuB) would be 0.5+0.5-0.25 = 0.75???
But isn't that saying that 75% are defective somehow and 25% are fine? How do you know? Isn't it possible that the 50% with defective handles are the exact same 50% with defective blades? Or perhaps every screwdriver is defective with either it having a defective blade or defective handle but no overlapping? I don't see how it makes sense.
It's possible, but in that case the two events (handles/blades) are hardly independent, are they?

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