[MOVED] Help Figuring out odds of 4 faulty products in a row

[keys360]

lol, I am but a simple Corrections Officer. Far from a math wiz.
My question is fairly simple (i think)

Say you buy a new product and the fail rate of that product was 3 out of 100 (3%). What would the odds of getting 4 faulty products in a row be?

 

[galactus]

I'd think it would depend on the size of the sample the items are selected from.

But as it sets, I'd say \(\displaystyle (\frac{3}{100})^{4}\approx{.00000081}\)

pka, one of the forum's leading probability experts , will hopefully be along to correct me if I'm wrong.

 

[pka]

Say you buy a new product and the fail rate of that product was 3 out of 100 (3%). What would the odds of getting 4 faulty products in a row be?

I'd think it would depend on the size of the sample the items are selected from.
But as it sets, I'd say \(\displaystyle (\frac{3}{100})^{4}\approx{.00000081}\)

It does indeed depend upon the size of a sample.
But let’s assume that we choose just four. Then Galactus correctly gave us the probability that all four are found to be defective.

However, the question asks for the odds all four are defective.

Odds in favor of \(\displaystyle A = \frac{{P(A)}}{{P(\bar A)}} = \frac{{P(A)}}{{1 - P(A)}}.\)

In this case: \(\displaystyle \frac{{\left( {0.03} \right)^4 }}{{1 - \left( {0.03} \right)^4 }} = \frac{{{\rm{541}}}}{{{\rm{667900410}}}}.\)

 

[galactus]

Hey pka. Thanks for the small correction, but isn't that very, very close to the same thing.

The difference between the two is only .000000000000656(there abouts). My bad. I thought when they said odds, it meant probability.

I see the difference now.

 

[pka]

The difference between the two is only .000000000000656(there abouts).

Yes indeed! But some people make a great deal out of this difference.
The truth is that I taught probability to undergraduates for 30 years, but I never said anything about 'odds'. Many will tell you that I did then a disservice, but I cannot see why. So you are right once again.

 

[keys360]

Wow. You guys are awsome. Please excuse me for being so bad at math. lol I was a B student in math, but only went as far as grade ten (Tornoto Canada).
Perhaps I should reword my question a bit? Honestley I have no clue. It's just your answer is not what I thought I was asking for. (lol, Although I'm sure I did get what I asked for)

Let me try again with a new question.

I am buying a product and the fail rate of that product is 3%. Their are 5 million of said product available for purchace. If I was to buy 4 of said product, what are the (aprox) odds that all 4 of said product would be defective?

The answer I'm looking for would be one that was broken down into somthing like... 1 in 100, 1 in a 1000000. The real answer being 1 in ??????

Again I am sorry if this all seems silly to you. I'm just that slow.
Please do help.

Thanks very much

 

[mcrae]

pka's answer of 541/667900410 giives you the % chance that all four are defective. 541/667900410= 0.000000810001%. in other words, 1/1234566.377 (if you buy 4 of these products 1234566 times, in one of those instances all four will be defective

 

[keys360]

pka's answer of 541/667900410 giives you the % chance that all four are defective. 541/667900410= 0.000000810001%. in other words, 1/1234566.377 (if you buy 4 of these products 1234566 times, in one of those instances all four will be defective

That is perfect. Way to break it down for the slow kid (lol me). Thank you so much.

 

[galactus]

Exactly..

This can be done with what is known as a hypergeometric distribution.

Assuming the sampling is done without replacement(that is, you don't return the defective item). Given a population of N items(in your case, N=5,000,000) having k successes(0 successes because they're all failures) and N-k failures(4 in this case), the probability of selecting a sample of size n(where n=4) that has x successes(0 successes) and n-x failures(4) is given by:

Since 3% of 5000000 is 150000.

\(\displaystyle \frac{C(k,x)C(N-k,n-x)}{C(N,n)}\)

\(\displaystyle \frac{C(4850000,0)C(150000,4)}{C(5000000,4)}\)

Here's the astonishing result I got:

\(\displaystyle \frac{21092906260312462500}{26041635416678124998750000}\)

Or about .00000081.

The difference between the result already obtained is not worth mentioning. I just rounded it off.

In other words, your odds are about 81 in 100 million of selecting 4 bad items in a row out of 5 million if 3% are rated as defective.

You'd have to have some mighty bad luck. It seems that way with me sometimes when I go shopping. Like that character in L'il Abner who went around with the rain cloud over his head all the time.

I know this probably looks like Greek to you, but it shows that probability can be an involved subject if you want to get particular.