# Easy discrete probability problems

And when I say easy I mean easy for a lot of you but not necessarily for most people.

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

The answer the book gives is $$\frac{_{90} C _{10}}{_{100} C _{10}}$$, which I think is wrong and it does not really make much sense. I think the correct answer is $$\frac{_{90}C_{4}}{_{100}C_{4}}$$.

2) Four microprocessors are randomly selected from a lot of 100 microprocessors among which 10 are defective. Find the probability of obtaining at most one defective microprocessor.

Well our sample space is still $$_{100}C_4$$ but the number of outcomes is different. If it was to be exactly one defective microprocessor then the number of outcomes would be $$_{10}C_1$$, the number of ways to select a defective microprocessor, times $$_{90}C_3$$, the number of ways to pick three nondefective processors. But when we say "at most one" we mean to say one or none. So can we just add the propability of getting none to the propability of getting one?

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xanthym
Townsend said:
And when I say easy I mean easy for a lot of you but not necessarily for most people.

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

The answer the book gives is $$\frac{_{90} C _{10}}{_{100} C _{10}}$$, which I think is wrong and it does not really make much sense. I think the correct answer is $$\frac{_{90}C_{4}}{_{100}C_{4}}$$.

2) Four microprocessors are randomly selected from a lot of 100 microprocessors among which 10 are defective. Find the probability of obtaining at most one defective microprocessor.

Well our sample space is still $$_{100}C_4$$ but the number of outcomes is different. If it was to be exactly one defective microprocessor then the number of outcomes would be $$_{10}C_1$$, the number of ways to select a defective microprocessor, times $$_{90}C_3$$, the number of ways to pick three nondefective processors. But when we say "at most one" we mean to say one or none. So can we just add the propability of getting none to the propability of getting one?
CORRECT for both your solutions. (Book is wrong on #1.)

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