09 Apr 08
3 edits
Originally posted by kbaumenunless this is a trick question, then the solution is as follows:
I'd like to post a problem also. I know that's not probability and is quite easy, but still I've heard some people answering wrongly.
Consider a bacteria in a sterile glass. Only the bacteria and the glass. Once in a minute, the bacteria reproduces and so the number of bacterias in the glass is doubled then. In an hour, the glass is full. Now two of these ...[text shortened]... kind of bacterias are placed in the glass. How long would it now take to have the glass full?
after 1 min 2^1 bacteria
after 2 min 2^2 bacteria
after 3 min 2^3 bacteria...
after 60 min 2^60 bacteria
for 2 bacteria
after 1 min 2^2 bacteria
after 2 min 2^3 bacteria
after 3 min 2^4 bacteria...
after 59 min 2^60 bacteria
so the answer is 59 minutes.
09 Apr 08
Originally posted by mtthwHere's another example:
And this, folks, is a good example as to why this conditional probability stuff matters.
Can a bunch of lawyers and jurors understand a conditional probability question? Get it wrong and an innocent person goes to prison...
http://www.badscience.net/?p=318
09 Apr 08
Originally posted by eldragonflyCorrect.
unless this is a trick question, then the solution is as follows:
after 1 min 2^1 bacteria
after 2 min 2^2 bacteria
after 3 min 2^3 bacteria...
after 60 min 2^60 bacteria
for 2 bacteria
after 1 min 2^2 bacteria
after 2 min 2^3 bacteria
after 3 min 2^4 bacteria...
after 59 min 2^60 bacteria
so the answer is 59 minutes.
10 Apr 08
Originally posted by PBE6Less than 1 in 50,000
Lol. 🙂
OK, one more. The rate of infection from a particular disease is 1 in 1,000,000 in the general population. A hospital wants to administer a test that is 99% accurate (i.e. if 100 people who have the disease get tested, 99 will test "positive" and 1 will test "negative" falsely) and 95% specific (i.e. if 100 who don't have the disease get tested, 95 ...[text shortened]... ts a positive result on the test, what is the chance that they really have the disease?
10 Apr 08
Originally posted by ThomasterCan't we discount that information?
no, it is 99 in 5.000.094
cause the test is only 99% accurate
1. We know that 1 in a million have the disease independently of any test.
2. Because our subject tested positive we know that he is not in the 1% that test negative when actually positive.
According to the information provided, if the 100 in 100 million that have the disease are tested one will show a negative result. This doesn't impact on our subject though as he/she tested positive.
10 Apr 08
1 edit
Originally posted by Green PaladinYes, I think I missed a step in the logic.
100 in 100 000 000 have the disease. Of the healthy 99 999 900 5% test positive when actually negative (4 999 995). Since the subject has tested positive he/she has to be in one of these two groups.
So the probability of a correct test result is 100 in 4 999 995 or 0.00002%?
Because the subject tested positive he/she must be one of the 99 out of the 100 out of the 100 million (or more likely the group of 4 999 995). He/she cannot be one of the hundred because one of them tested negative which we know is not the case with our subject.
So the probability of a correct test result is 99 in 4 999 995 (1 in 50505) or 0.0000198% ?
10 Apr 08
Originally posted by Green PaladinIt is 99 in 5.000.094
Yes, I think I missed a step in the logic.
Because the subject tested positive he/she must be one of the 99 out of the 100 out of the 100 million (or more likely the group of 4 999 995). He/she cannot be one of the hundred because one of them tested negative which we know is not the case with our subject.
So the probability of a correct test result is 99 in 4 999 995 or 0.0000198%?
99/5.000.094x100=0,001979962777%
This seems to be a useless test
10 Apr 08
1 edit
Originally posted by Green Paladin1 in 50506. You must be missing one small step.
So the probability of a correct test result is 99 in 4 999 995 (1 in 50505) or 0.0000198% ?
Remember the sample space is all those that would test positive. So it would be:
99/(99 + 4999995)