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Mathematics 4 Online
OpenStudy (anonymous):

A shelf in the Metro Department Store contains 95 colored ink cartridges for a popular ink-jet printer. Five of the cartridges are defective. (a) If a customer selects 3 cartridges at random from the shelf, what is the probability that they are all defective? (Round your answer to five decimal places.) (b) If a customer selects 3 cartridges at random from the shelf, what is the probability that at least 1 is defective? (Round your answer to three decimal places.)

hartnn (hartnn):

from 95, 5 are defective, so what is the probability of finding 1 defective cartridge ?

OpenStudy (anonymous):

5/95, so 1/19

hartnn (hartnn):

right, now we need to find that all 3 cartridge are defective, so first lets find when 2 are chosen and both are defective, what is the probability for this? any ides ?

OpenStudy (anonymous):

(5/95)*(4/94) maybe?

hartnn (hartnn):

absolutely correct! so now, in same way, what about 3?

OpenStudy (anonymous):

(5/95)*(4/94)*(3/93) so 0.000072?

hartnn (hartnn):

yup.

hartnn (hartnn):

to 5 decimal places it would only be 0.00007.

OpenStudy (anonymous):

Awesome. The second part of the question is actually pick 2 not pick 3, and at least one is defective. Would that be (5/95) * (1/94) ?

hartnn (hartnn):

had it been 3, it would have been really easy. to get the probability of atleast 1 defective, you first find the probability that none of them are defective.

OpenStudy (anonymous):

None are defective is 90/95, defective is 5/95

hartnn (hartnn):

i meant none of the 2 are defective, so 90/95 multiplied by what ?

OpenStudy (anonymous):

(90/95)*(89/94)

hartnn (hartnn):

correct, now use this P(atleast 1) = 1- P(none)

hartnn (hartnn):

so 1- product of those 2 will give u probability of atleast 1

hartnn (hartnn):

u there ? did u understand that ?

OpenStudy (anonymous):

A little, I'm trying to figure it out and write it out but I'm getting confused. Is it 1-(5/95)/(90/95)?

hartnn (hartnn):

nopes, its 1- (90/95)*(89/94). just use this when u want to find the probability of ATLEAST 1. P(ATLEAST 1) = 1- P(NONE)

OpenStudy (anonymous):

Oh! Ok so it's..... one sec I'm going to bust out my calculator

hartnn (hartnn):

me too :P

OpenStudy (anonymous):

I'm doing this completely wrong, I'm sorry.

hartnn (hartnn):

why? what happened ?

OpenStudy (anonymous):

Trying to find pick two at least one is defective. So for the one that IS defective it is 5/95 and for the second one that can be either defective or normal, would that be 1/94? I don't understand why there is (90/95)*(89/94). Shouldn't it just be (5/95)*(1/94)?

OpenStudy (anonymous):

But that isn't right and I don't understand why.

hartnn (hartnn):

just a sec, let me think...

hartnn (hartnn):

'for the second one that can be either defective or normal, would that be 1/94'<--NO we are first finding the probability that when 2 are selected, BOTH of them are NOT defective. what is that probability ?

OpenStudy (anonymous):

90/95 * 89/94.

hartnn (hartnn):

u clear with this, right ? now to find the probability that atleast one of the cartridge is defective, we do 1-probability that both (all) the cartridges are NOT defective.

OpenStudy (anonymous):

Ok, so that would be 0.10302. I can understand that sort of.

hartnn (hartnn):

yeah, thats correct.

hartnn (hartnn):

just remember this : P(ATLEAST 1) = 1- P(NONE)

OpenStudy (anonymous):

so that would be 5/95 = 1-90/95 * 89/94 ?

OpenStudy (anonymous):

Oh nevermind, that is just x = 1-90/95 * 89/94

OpenStudy (anonymous):

Yes!! That's right! Thank you so much!!

hartnn (hartnn):

welcome ^_^ and \(\huge \color{red}{\text{Welcome to Open Study}}\ddot\smile\)

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