Simple probability question that I want to have work checked

[thepatient]

Homework Statement

A manufacturing machine produces defects with a probability of 0.1%. How many parts must the machine produce to have a 99.9% chance of producing at least 1 defective part?

Homework Equations

P(A) + P(B) = 1

The Attempt at a Solution

[/B]
A in this case is the machine produces at least one defective part
B is the case when the machine produces all good parts

The probability that the machine produces a good part in the first try is:
P(B) = 1 - .001 = .999

The probability that the machine produces two good parts consecutively is:
P(B) = .999*.999 = .999^2

So I assume the probability that the machine produces n good parts consecutively would be:
P(B) = .999^n

Therefore the probability that the machine produces at least 1 bad part part within n consecutive parts must be:
P(A) = 1 - .999^n

Using P(A) = .999 and solving for n.

.999 = 1 - .999^n

ln(.001)/(ln(.999) = n = 6904.3 = 6905 parts

Does that make sense? That we need to produce 6905 parts to have a 99.9 percent chance of 1 defective part?

 

[Ray Vickson]

Homework Statement

A manufacturing machine produces defects with a probability of 0.1%. How many parts must the machine produce to have a 99.9% chance of producing at least 1 defective part?

Homework Equations

P(A) + P(B) = 1

The Attempt at a Solution

A in this case is the machine produces at least one defective part
B is the case when the machine produces all good parts

The probability that the machine produces a good part in the first try is:
P(B) = 1 - .001 = .999

The probability that the machine produces two good parts consecutively is:
P(B) = .999*.999 = .999^2

So I assume the probability that the machine produces n good parts consecutively would be:
P(B) = .999^n

Therefore the probability that the machine produces at least 1 bad part part after n consecutive parts must be:
P(A) = 1 - .999^n

Using P(A) = .999 and solving for n.

.999 = 1 - .999^n

ln(.001)/(ln(.999) = n = 6904.3 = 6905 parts

Does that make sense? That we need to produce 6905 parts to have a 99.9 percent chance of 1 defective part?[/B]

The probability that all of the first $n$ parts are good is $P_n = 0.999^n$, so the complement of that is the probability that not all of the first $n$ parts are good. Your use of the word "after" threw me; I would rather use the word "among" or "within", because that is exactly what the quantity $1-P_n$ would mean.

Anyway, you computations make sense and are also correct.

PS: why do you write in all bold font? It makes it look like you are shouting!

 

[thepatient]

Oops sorry. I clicked the space after the 3rd question, and everything after that came out bold. I just realized now that there is a formatting tool above. I'll edit that.

 

[thepatient]

Thanks for taking a look.