conditional probability or geometric distribution

[mcrae]

80 good and 10 bad screws are in a box. if 10 screws are used what iss the probability that no screws are bad?

i did (80/90)^10 = 0.307946

teacher did [(10C0)(80C10)]/90C10 = 0.287815

he said i might be right but never followed up on it cause hes a moron, so can anyone tell me which answer is more correct and why the other one is wrong or a case where the two methods would yeild extremely different answers?

 

[tkhunny]

Your version can't be exact. You don't put the screws back after you use them, do you? As you remove good screws, the probability of getting a bad one increases. This makes it a little more difficult than your formulation has predicted. That the teacher's answer is just a little lower than yours is the good news.

Getting along with your teacher usually is a good idea.

I wouldn't call those "extremely different answers". It looks almost like a reasonable approximation, particularly if you know the direction of the error. Of course, if you were running a gambling house, you'ld need to be more precise.

Doing it the hard way, one sees that:

\(\displaystyle \frac{80*79*78*77*76*75*74*73*72*71}{90*89*88*87*86*85*84*83*82*81}\,=\,0.287815792\)

Now what do you think of your teacher's answer?

 

[mcrae]

oh, right. thanks for that then

my teacher is a moron cause he says he'll explain things to me and then never does... but then i come online and everything is good

 

[Denis]

my teacher is a moron cause he says he'll explain things to me and then never does...

That doesn't mean he's a moron; perhaps you have bad breath or something :shock:

 

[soroban]

Hello, mcrae!

80 good and 10 bad screws are in a box.
If 10 screws are used, what is the probability that no screws are bad?

i did (80/90)^10 = 0.307946

teacher did [(10C0)(80C10)]/90C10 = 0.287815

he said i might be right but never followed up on it cause hes a moron

He's a moron all right . . . He should have told you immediately that you are wrong.
His answer is correct.

There are 90 screws and 10 are chosen.
There are: \(\displaystyle _{90}C_{10}\,=\,\frac{90!}{10!80!}\) ways to do this.

There are 80 good screws and 10 are chosen.
There are: \(\displaystyle _{80}C_{10}\,=\,\frac{80!}{10!70!}\) ways to do this.

Therefore: \(\displaystyle \;P(\tex{10 good})\:=\:\frac{_{90}C_{10}}{_{80}C_{10}}\:=\:0.287815792...\)


Your method assumes that every time you reach in the box,
\(\displaystyle \;\;\)there are 90 screws (and 80 of them are good).