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# Calling all stats gurus. Can you solve this? watch

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1. Hi,

I am looking for somebody with a stats background who can solve the following probability problem:

You have a box. It contains 30 computer disks. 14 are blank and 16 are not blank. What is the probability that you must draw at least one diskette before finding a blank one?

2. um

mayb you've written the question wrong but from what you've written

you want the prob that you have to select a disk before you select a blank disk

well obviously the probability is 1.

you cant draw a blank disk before you draw a disk.

love Katy ***
3. "You have a box. It contains 30 computer disks. 14 are blank and 16 are not blank. What is the probability that you must draw at least one diskette before finding a blank one?"

OK, so the question is basically saying that the first diskette drawn must be NOT BLANK. After that it doesnt matter.

so....the probability is: 1 - P(1st diskette IS NOT BLANK)
= 1 - 16/30 = 14/30
4. (Original post by Unregistered)
"You have a box. It contains 30 computer disks. 14 are blank and 16 are not blank. What is the probability that you must draw at least one diskette before finding a blank one?"

OK, so the question is basically saying that the first diskette drawn must be NOT BLANK. After that it doesnt matter.

so....the probability is: 1 - P(1st diskette IS NOT BLANK)
= 1 - 16/30 = 14/30
What if the second diskette is not blank?

What is the question meant to be, do we replace after each drawing?
5. (Original post by theone)
What if the second diskette is not blank?

What is the question meant to be, do we replace after each drawing?
it doesnt matter if the second is blank, the question just states 'at least one', so its any possiblily except the first being not blank, hence it being 1 - P(1st not blank)
6. I agree, but I was actually hinting at a more technical method called the negative binomial expansion (only works with replacement) from which you can also find out the expected number of non-blank diskettes drawn and the variance.
7. (Original post by theone)
I agree, but I was actually hinting at a more technical method called the negative binomial expansion (only works with replacement) from which you can also find out the expected number of non-blank diskettes drawn and the variance.
hehe, yeh fair enough if its replaced, I follow, stats 1 is a distant memory now
8. Just out of interest, since this example is more relevant to negative bionmial, what is the probability that if i roll a fair die my 3rd 6 is on my 7th roll?

This is like this except we're only considering the first blank diskette which makes the whole problem a lot simpler....
9. I'm off to bed so i'll post the answer.

We need a 6 on our 7th roll, with p = 1/6 and 2 sixes from 6 rolls in our previous 6 rolls = (1/6)^2 . (5/6)^4 . 6C2.

Multiply the two together et voila.

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