Probability Question I am stuck on - Tilted Forum Project Discussion Community

Y2KDREAD · 10-30-2004, 07:49 PM

I know this should be a really easy question I should be able to get in two questions, but I suck at math. We are talking about combonations and permitations, so use that crap.

From among 80 light bulbs, 10 are defective. How man samples of 8 have at least 1 deffective bulb.

Can anyone help?

molloby · 10-30-2004, 08:20 PM

All right. X~B(8,1/8)

P(X/=0) = 1-P(X=0) = 1-(7/8)^8 = 0.656391

Therefore from 10 samples the expected number with at least one defective bulb is 6.56391 or 7.

Y2KDREAD · 10-30-2004, 08:27 PM

i think that is way too complicated for my calss, just checked on the profs website and she posted the answer, she told it is 19,547,186,230

molloby · 10-30-2004, 08:42 PM

Where is she pulling the numbers from, how may samples are we talking?

Getting 19.5 billion samples of 8 from a set of 80 is quite impressive...

Y2KDREAD · 10-30-2004, 08:46 PM

i have no idea, all i know is that when we figure it out in the calculator (TI-83+ for me), we are using the nCr and and nPr things. I assume this one will be nCr because order does not matter. I figured that it would be 80 nCr 10 or (C(80,10)) divided or subtracted by something that represnts the C(8,1), i just can't figure it out.

molloby · 10-30-2004, 09:08 PM

I think I've got it:

There are 80C8 unique ways of choosing 8 balls from 80 and 70C8 ways of choosing 8 balls from 70. So:

Number of with defects = Total number of choices - number of choices without defects

= 80C8-70C8
=19,547,718,623

Enjoy.

Y2KDREAD · 10-30-2004, 09:14 PM

thanks a lot, i don't really have a math sorta mind (i can do it, but it takes me awhile) but it takes me a bit to figure stuff out, and half the time i get fed up and just skip stuff.

molloby · 10-30-2004, 09:19 PM

Fair enough.

Honestly I'm just avoiding my own work, it is always easier to do someone else's than your own.

Y2KDREAD · 10-30-2004, 09:22 PM

and it gives a sense of accomplishment that you don't get from doing your own work

fckm · 10-31-2004, 11:30 AM

No offense, but it seems as if your teacher's not doing her job correctly. She should be teaching you how to do probability and statistics, not how to plug numbers in to you calculator. I would suggest that you go back to your textbook and reread the section that defines what permutations and combinations are. Hopefully, you'll gain a better understanding of why they are defined the way they are and how they work.

10-30-2004, 07:49 PM	#1 (permalink)
Y2KDREAD Crazy Location: College Station, TX	Probability Question I am stuck on I know this should be a really easy question I should be able to get in two questions, but I suck at math. We are talking about combonations and permitations, so use that crap. From among 80 light bulbs, 10 are defective. How man samples of 8 have at least 1 deffective bulb. Can anyone help? __________________ Signatures are for chumps.

10-30-2004, 08:27 PM	#3 (permalink)
Y2KDREAD Crazy Location: College Station, TX	i think that is way too complicated for my calss, just checked on the profs website and she posted the answer, she told it is 19,547,186,230 __________________ Signatures are for chumps.

10-30-2004, 08:46 PM	#5 (permalink)
Y2KDREAD Crazy Location: College Station, TX	i have no idea, all i know is that when we figure it out in the calculator (TI-83+ for me), we are using the nCr and and nPr things. I assume this one will be nCr because order does not matter. I figured that it would be 80 nCr 10 or (C(80,10)) divided or subtracted by something that represnts the C(8,1), i just can't figure it out. __________________ Signatures are for chumps.

10-30-2004, 09:14 PM	#7 (permalink)
Y2KDREAD Crazy Location: College Station, TX	thanks a lot, i don't really have a math sorta mind (i can do it, but it takes me awhile) but it takes me a bit to figure stuff out, and half the time i get fed up and just skip stuff. __________________ Signatures are for chumps.

10-30-2004, 09:22 PM	#9 (permalink)
Y2KDREAD Crazy Location: College Station, TX	and it gives a sense of accomplishment that you don't get from doing your own work __________________ Signatures are for chumps.

10-30-2004, 08:20 PM	#2 (permalink)
molloby Tilted Location: Sydney, Australia	All right. X~B(8,1/8) P(X/=0) = 1-P(X=0) = 1-(7/8)^8 = 0.656391 Therefore from 10 samples the expected number with at least one defective bulb is 6.56391 or 7.

10-30-2004, 08:42 PM	#4 (permalink)
molloby Tilted Location: Sydney, Australia	Where is she pulling the numbers from, how may samples are we talking? Getting 19.5 billion samples of 8 from a set of 80 is quite impressive...

10-30-2004, 09:08 PM	#6 (permalink)
molloby Tilted Location: Sydney, Australia	I think I've got it: There are 80C8 unique ways of choosing 8 balls from 80 and 70C8 ways of choosing 8 balls from 70. So: Number of with defects = Total number of choices - number of choices without defects = 80C8-70C8 =19,547,718,623 Enjoy.

10-30-2004, 09:19 PM	#8 (permalink)
molloby Tilted Location: Sydney, Australia	Fair enough. Honestly I'm just avoiding my own work, it is always easier to do someone else's than your own.

10-31-2004, 11:30 AM	#10 (permalink)
fckm Insane Location: Ithaca, New York	No offense, but it seems as if your teacher's not doing her job correctly. She should be teaching you how to do probability and statistics, not how to plug numbers in to you calculator. I would suggest that you go back to your textbook and reread the section that defines what permutations and combinations are. Hopefully, you'll gain a better understanding of why they are defined the way they are and how they work.