What is the correct probability for obtaining no defective microprocessors?

[Townsend]

There is a question in my text that list an answer in the back of the book that seems wrong to me.
It is,

"Four microprocessors are randomly selected from a lot of 100 microprocessors among which 10 are defective. Find the probability of obtaining no defective microprocessors."

There are C(100,4) ways of selecting 4 microprocessors from 100 and this is our sample space. There are 90 non-defective microprocessors and so there are C(90,4) ways to select a non-defective microprocessor. So our number of out comes of the event is C(90,4) and so the probability of the event is

P(E)=C(90,4)/C(100,4).

The answer in the back of the textbook is given as

C(90,10)/C(100,10)

This answer does not make much sense to me since we are selecting 4 things and not 10 things. Am I right or is the book right?

Thanks

 

[Gokul43201]

The book is incorrect; you are not.

 

[Townsend]

Thanks Gokul...

I have another quick question,

"Four microprocessors are randomly selected from a lot of 100 microprocessors among which 10 are defective. Find the probability of obtaining exactly one defective microprocessor"

The sample space is C(100,4) but the outcomes of the event are a bit different. Here there are C(90,4) ways to get all good processors but we want exactly one bad one in our 4. Well the other 3 must be good so they are C(90,3) ways for that to happen and there are C(10,1) ways to get one bad processor. So I am thinking that the ways to get the good processors plus the ways to get the bad processor make up the total number of ways to get exactly one bad processor. So

P(E)=(C(10,1)+C(90,3))/C(100,4)

is the probability of obtaining exactly one bad microprocessor.

Thanks

 

[learningphysics]

Thanks Gokul...

I have another quick question,

"Four microprocessors are randomly selected from a lot of 100 microprocessors among which 10 are defective. Find the probability of obtaining exactly one defective microprocessor"

The sample space is C(100,4) but the outcomes of the event are a bit different. Here there are C(90,4) ways to get all good processors but we want exactly one bad one in our 4. Well the other 3 must be good so they are C(90,3) ways for that to happen and there are C(10,1) ways to get one bad processor. So I am thinking that the ways to get the good processors plus the ways to get the bad processor make up the total number of ways to get exactly one bad processor. So

P(E)=(C(10,1)+C(90,3))/C(100,4)

is the probability of obtaining exactly one bad microprocessor.

Thanks

You have to get 1 bad AND 3 good... so you have to multiply:
P(E)=(C(10,1)*C(90,3))/C(100,4)

 

[Gokul43201]

Yes, these are not mutually exclusive events. They are independent, sequential events.

There are 10 ways of picking the defect. For each of these 10 choices, there are C(90,3) ways of picking the 3 good ones. So the "total" here will be the product.