|
|
Re: Conditional Probability Question
Posted:
Sep 25, 2001 2:33 PM
|
|
"Aaron Davies" <aaron@avalon.pascal-central.com> wrote in message
news://1f0aj51.1aa72ty1r5nx88N%aaron@avalon.pascal-central.com...
> There is a bucket containing 100 machine parts, one of which is known to
> be defective. You draw parts from the bucket one at a time, examining
> them for the defect, and then set them aside (ie no replacement). What
> is the probability that the i-th part you examine will be defective?
>
> The professor is of the opinion that the probability is always 1/100 and
> is thus independant of i. Most of the students are of the opinion that
> it is 1/(101-i). Which is correct?
The answer of "most of the students" can immediately be seen to be wrong,
because the sum of the probabilities is greater than 1.
The probability that the first piece you pick up is defective is, everyone
agrees, 1/100. The probability that it is OK, is 99/100.
The probability that you inspect a second piece is therefore 99/100. The
probability that you inspect it and find it to be defective is
(99/100)*(1/99) = 1/100. [This is perhaps the crucial realization. Be sure
you understand that.] The probability that you inspect it and find it to be
OK is (99/100)*(98/99) = 98/100.
The probability that you inspect a third piece, therefore, is 98/100. The
probability that you inspect it and find it to be defective is
(98/100)*(1/98) = 1/100.
And so on.
I hope this helps you.
|
|
