conditional probability or geometric distribution

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?

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?

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

mcrae said:

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

Click to expand...

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

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

Click to expand...

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).