View Full Version : 10$ if you can do this in the next 20 minutes and explain it to me
If you can do this via a tree (using paint I suppose) I'll transfer you $10. I'm about to take a test with a question like this on it and this one has me stumped. I know it's stupidly simple but I can't seem to organize it right in my head.
It is known that 20 % of a school's athlete population uses drugs. The Athletic Director administers a drug test for which a positive result is correct 96 % of the time, and a negative result is correct 92 % of the time. One athlete is randomly selected and tested. Find the probability that the athlete is a drug user, given that the test result is positive. Give your answer as a decimal number, with three decimal places.
$15 and you have 25 minutes.
MDoranD
03-11-2006, 10:34 AM
96%???
diebitter
03-11-2006, 10:35 AM
0.96 isn't it?
If the a priori assumption is he is a drug user, and he tests +ve, then it's the reliability of the test's prob.
No it's much less than that.
MDoranD
03-11-2006, 10:37 AM
20% of the athletes using drugs shouldnt matter if you state that the test reslut was positive and its 96% accurate.
The answer is 0.75±0.01
I need to know how to set the tree up to get there.
Here's another:
It is known that 10 % of a school's athlete population uses drugs. The Athletic Director administers a drug test for which a positive result is correct 90 % of the time, and a negative result is correct 92 % of the time. One athlete is randomly selected and tested. Find the probability that the athlete is a drug user, given that the test result is positive. Give your answer as a decimal number, with three decimal places.
Answer: 0.556±0.01
Matt R.
03-11-2006, 11:06 AM
Hi,
It's a conditional probability question.
Let F be the event that an athlete tests positive.... P(F) = (.96*.2) + (.08*.8). (Quick edit to show you where I got the 2nd term in the sum)
Let E be the event that he is a drug user. So P(EF) is .2*.96
So the prob of E given F is P(E|F) = P(EF)/P(F) = .2*.96 / ((.96*.2) + .064) = .75.
Matt R.
03-11-2006, 11:07 AM
I noticed you were short on time so I typed it up as quickly as possible -- if you're still there and need a further explanation for anything let me know.
diebitter
03-11-2006, 12:50 PM
Ay! You changed the original posit. You said he 'was a drug user' or similar originally, and now you've changed it. Don't make us look bad!
Thanks- I didn't get your message in time, but luckily I managed through it come test time by making a Venn diagram. Scored 100% too. =]
diebitter
03-11-2006, 02:51 PM
[ QUOTE ]
I never edited anything.
[/ QUOTE ]
In that case, sorry. I could have shown you said he was a drug user. My bad.
madnak
03-12-2006, 06:20 PM
[ QUOTE ]
In that case, sorry. I could have shown you said he was a drug user. My bad.
[/ QUOTE ]
Tough day?
The question is improperly worded. If a positive result is correct 96% of the time, and the athlete tests positive, then there is a 96% chance he is a drug user.
What you mean is that if he is a drug user, there is a 96% chance of a positive and a 4% chance of a negative result. But if he's not a drug user, there is a 92% chance of a negative and an 8% chance of a positive result.
Therefore, if 20% of students are drug users, only ~75% of positive results will actually be accurate.
Borodog
03-12-2006, 08:53 PM
[ QUOTE ]
[ QUOTE ]
In that case, sorry. I could have shown you said he was a drug user. My bad.
[/ QUOTE ]
Tough day?
The question is improperly worded. If a positive result is correct 96% of the time, and the athlete tests positive, then there is a 96% chance he is a drug user.
What you mean is that if he is a drug user, there is a 96% chance of a positive and a 4% chance of a negative result. But if he's not a drug user, there is a 92% chance of a negative and an 8% chance of a positive result.
Therefore, if 20% of students are drug users, only ~75% of positive results will actually be accurate.
[/ QUOTE ]
Ding ding ding.
stinkypete
03-12-2006, 10:02 PM
[ QUOTE ]
[ QUOTE ]
In that case, sorry. I could have shown you said he was a drug user. My bad.
[/ QUOTE ]
Tough day?
The question is improperly worded. If a positive result is correct 96% of the time, and the athlete tests positive, then there is a 96% chance he is a drug user.
What you mean is that if he is a drug user, there is a 96% chance of a positive and a 4% chance of a negative result. But if he's not a drug user, there is a 92% chance of a negative and an 8% chance of a positive result.
Therefore, if 20% of students are drug users, only ~75% of positive results will actually be accurate.
[/ QUOTE ]
this is the correct answer.
Marko Schmarko
03-13-2006, 12:30 AM
[ QUOTE ]
Therefore, if 20% of students are drug users, only ~75% of positive results will actually be accurate.
[/ QUOTE ]
This seems wrong to me.
It seems to me that the percentage of drug users with a positive result would be .2*.96 and the percentage of non-users with positive results would be .8*.04 (I can't imagine why the percentage of false-negatives would be relevant). This would represent the entire population of positive results.
Thus, the percentage of drug users of the total positive population would be (.2*.96)/(.2*.96+.8*.04) ~= .86.
Where have I gone wrong?
diebitter
03-13-2006, 02:39 AM
[ QUOTE ]
[ QUOTE ]
Therefore, if 20% of students are drug users, only ~75% of positive results will actually be accurate.
[/ QUOTE ]
This seems wrong to me.
It seems to me that the percentage of drug users with a positive result would be .2*.96 and the percentage of non-users with positive results would be .8*.04 (I can't imagine why the percentage of false-negatives would be relevant). This would represent the entire population of positive results.
Thus, the percentage of drug users of the total positive population would be (.2*.96)/(.2*.96+.8*.04) ~= .86.
Where have I gone wrong?
[/ QUOTE ]
You havevn't read a textbook on Bayesian probability.
Marko Schmarko
03-13-2006, 03:17 AM
What an utterly useless response.
Sincerely,
Mark
tomdemaine
03-13-2006, 04:00 AM
and yet accurate in every respect.
Marko Schmarko
03-13-2006, 04:33 AM
[ QUOTE ]
Let F be the event that an athlete tests positive.... P(F) = (.96*.2) + (.08*.8). (Quick edit to show you where I got the 2nd term in the sum)
[/ QUOTE ]
This is the part that doesn't make sense to me.
Wouldn't P(F) be (.96*.2 + .04*.8)? Wouldn't the .08*.8 be negative-testing drug users, while .04*.8 be positive-testing straight-edge kids?
Could someone clarify?
Marko Schmarko
03-13-2006, 04:44 AM
Hmm.. to make all the possibilities add to 1 in a logical way, you have to use .96 * .2 + .08*.8 + .92* .8 + .04 * .2..
That clues me into at least a way to see that it has to be as you all have posited, but not how it makes sense that .08*.8 has any place within the +ve response event.
Marko Schmarko
03-13-2006, 05:09 AM
Hmm... Now I feel like an idiot, but I can't erase my second-to-last post.
.08*.8 would be false-negative testing non-drug users. Which... is obviously illogical. Does this mean that this is actually positive testing non-drug users?
This leap doesn't make sense to me.
Still searching for enlightenment.
- mark
diebitter
03-13-2006, 05:27 AM
I wouldn't feel an idiot about it. Asking questions to gain understanding, no matter how dumb they may sound, is not the sign of an idiot.
Double Down
03-13-2006, 09:52 AM
[ QUOTE ]
The Athletic Director administers a drug test for which a positive result is correct 96 % of the time,
Find the probability that the athlete is a drug user, given that the test result is positive.
[/ QUOTE ]
It's 96%, people. None of that other info matters.
The question is: If the test result is positive, what is the probability that he's a user? In the setup, it says 96%.
Duh
All that other info is there just to throw you off.
madnak
03-13-2006, 09:55 AM
[ QUOTE ]
This seems wrong to me.
It seems to me that the percentage of drug users with a positive result would be .2*.96 and the percentage of non-users with positive results would be .8*.04 (I can't imagine why the percentage of false-negatives would be relevant). This would represent the entire population of positive results.
Thus, the percentage of drug users of the total positive population would be (.2*.96)/(.2*.96+.8*.04) ~= .86.
Where have I gone wrong?
[/ QUOTE ]
Here's what you're missing. The percentage of nonusers with correct results is 92%. Therefore the percentage of nonusers with positive results is 8%, not 4%. 4% is the number of drug users who get negative results.
Thus, the percentage of drug users of the total positive population would be (.2*.96)/(.2*.96+.8*.08) = .75.
Double Down
03-13-2006, 10:08 AM
[ QUOTE ]
Thus, the percentage of drug users of the total positive population would be (.2*.96)/(.2*.96+.8*.04) ~= .86.
Where have I gone wrong?
[/ QUOTE ]
Guys, reread the question. If A player is tested and found to be positive, what are the chances of him actually being positive? 96%!
Copernicus
03-13-2006, 12:26 PM
[ QUOTE ]
[ QUOTE ]
Thus, the percentage of drug users of the total positive population would be (.2*.96)/(.2*.96+.8*.04) ~= .86.
Where have I gone wrong?
[/ QUOTE ]
Guys, reread the question. If A player is tested and found to be positive, what are the chances of him actually being positive? 96%!
[/ QUOTE ]
wrong double down, 75% is the correct answer. Your answer is to a different question: If a known user takes the test what is the probability it will show up positive? This asks about false negatives.
That is not the same as: If a test shows up positive what is the probability he is a user? This asks about false positives.
roundmound
03-13-2006, 01:12 PM
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Thus, the percentage of drug users of the total positive population would be (.2*.96)/(.2*.96+.8*.04) ~= .86.
Where have I gone wrong?
[/ QUOTE ]
Guys, reread the question. If A player is tested and found to be positive, what are the chances of him actually being positive? 96%!
[/ QUOTE ]
wrong double down, 75% is the correct answer. Your answer is to a different question: If a known user takes the test what is the probability it will show up positive? This asks about false negatives.
That is not the same as: If a test shows up positive what is the probability he is a user? This asks about false positives.
[/ QUOTE ]
This is exactly what the question asked. The question was worded wrong but most people in the thread answered assuming it was a prob question, which it was, and not a trick question.
DMBFan23
03-13-2006, 02:11 PM
[ QUOTE ]
If you can do this via a tree (using paint I suppose) I'll transfer you $10. I'm about to take a test with a question like this on it and this one has me stumped. I know it's stupidly simple but I can't seem to organize it right in my head.
It is known that 20 % of a school's athlete population uses drugs. The Athletic Director administers a drug test for which a positive result is correct 96 % of the time, and a negative result is correct 92 % of the time. One athlete is randomly selected and tested. Find the probability that the athlete is a drug user, given that the test result is positive. Give your answer as a decimal number, with three decimal places.
[/ QUOTE ]
reading the problem, it seems weirdly stipulated. A negative result being incorrect shouldn't mean that someone who ISN'T a user tests positive instead, it should mean that a user will obtain a negative result 8% of the time. conversely, a positive result being false shouldnt mean that users get negatives instead, it should mean that non-users get positives when they shouldnt. But under those interpretations, the probabilities don't add up to 1.
How am I misinterpreting?
mindflayer
03-13-2006, 02:13 PM
if you know the person IS a drug user then the prob that the person is a user given the result is positive is 100%
hahah
if you dont know the person is... (and he/she was randomly selected) then the venn diagram looks like this.
20% IS a Drug user x 96% positive = 19.20% positive result
20% IS a Drug user x 4% negative = 0.80% negative result
80% Is NOT a user x 8 % positive = 6.4% positive result
80% Is NOT a user x 92% negative = 73.6% negative result
randomly selected person will show 25.6% postive
74.4% negative
total prob 100.00%
prob the person is a user given the result was positive??
= ratio of positive users to positive total results
19.20% / 25.6% = 75%
Madnak 1 ; Double Down 0
Colima420
03-15-2006, 09:31 PM
The correct answer is 0.750. If you want me to send you the tree diagram. let me know. Basically the probability of the athlete being a drug user if the test result is positive is as follows: (Prob. athlete is a user * Prob. of positive test being correct) / (Prob. athlete is a user * Prob. of positive test being correct + Prob. of student not using drugs * Prob. of test giving a false positive). In numbers: [(0.96)*(.20)]/[(0.96)*(0.20)+(0.8)*(0.08)]=0.750
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