Easy discrete probability problems

[Townsend]

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?

 

[xanthym]

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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[thepatientmental]

1) The answer given in the book is actually correct. The formula used is for the probability of selecting a specific number of objects from a larger set, without replacement. In this case, we are looking for the probability of selecting 4 non-defective microprocessors from a lot of 100, which is represented by 90C4/100C4. This is because the number of ways to select 4 non-defective microprocessors is the same as the number of ways to select 4 objects from a set of 90 (the non-defective microprocessors).

2) The correct approach for this problem would be to calculate the probability of selecting exactly one defective microprocessor and adding it to the probability of selecting none. So the final answer would be:

P(at most one defective) = P(exactly one defective) + P(none defective)
= (10C1 * 90C3)/100C4 + (90C4)/100C4
= (10 * 11780)/3921225 + 118755/3921225
= 0.3114 + 0.0303
= 0.3417

Therefore, the probability of selecting at most one defective microprocessor is 0.3417.