# Thread: Probability of A given B is true

1. ## Probability of A given B is true

I'm pretty sure that's what kind of question this is anyway.

A drugs test for athletes is 95% effective in detecting a steroid wen it is actually present. However, it also yields a false positive result for 1% of "clean" athletes tested. If 0.5% of athletes use the steroid, what is the probability that an athlete who tests positive did actually use the drug?

2. Originally Posted by chella182
I'm pretty sure that's what kind of question this is anyway.

A drugs test for athletes is 95% effective in detecting a steroid wen it is actually present. However, it also yields a false positive result for 1% of "clean" athletes tested. If 0.5% of athletes use the steroid, what is the probability that an athlete who tests positive did actually use the drug?
Imagine that 100000 atheletes tested so that 0.005(100000)= 500 use the steroid and 99500 don't. Of the 500 who use the steroid 0.95(500)= 465 test positive. Of the 99500 who don't use the steroid 0.01(99500)= 995 also test positive. That gives a total of 995+ 465= 1465 positive tests of whom 465 actually use steroids. The probability that a positive test means the athelete did use the steroid is 465/1465.

3. Hello, chella182!

Yes, this is a Conditional Probability problem.

A drugs test for athletes is 95% effective in detecting a steroid when it is actually present.
However, it also yields a false positive result for 1% of "clean" athletes tested.
If 0.5% of athletes use the steroid, what is the probability that an athlete
who tests positive did actually use the drug?
"0.5% of the atheletes use the drug": .$\displaystyle P(\text{use}) \:=\:0.005,\;P(\text{clean}) \:=\:0.995$

"The test is 95% effective in detecting a steroid when it is actually present."
If the athlete Uses the drug, the test will be Positive 95% of the time.
. . That is: .$\displaystyle P(\text{pos}\,|\,\text{use}) \:=\:0.95$
From Bayes' Theorem: .$\displaystyle P(\text{pos}|\text{use}) \:=\:\frac{P(\text{pos}\wedge\text{use})}{P(\text{ use})} \:=\:0.95$
. . Then: .$\displaystyle \frac{P(\text{pos}\wedge\text{use})}{0.005} \:=\:0.95 \quad\Rightarrow\quad P(\text{pos}\wedge\text{use}) \:=\:0.00475\;\;{\color{blue}[1]}$

"It yields a false positive result for 1% of 'clean' athletes tested."
If an athlete is clean, the test will be positive 1% of the time.
. . That is: .$\displaystyle P(\text{pos}|\text{clean}) \:=\:0.01$
From Bayes' Theorem: .$\displaystyle P(\text{pos}|\text{clean}) \:=\:\frac{P(\text{pos}\wedge\text{clean})}{P(\tex t{clean})} \:=\:0.01$

. . Then: .$\displaystyle \frac{P(\text{pos}\wedge\text{clean})}{0.995} \:=\:0.01 \quad\Rightarrow\quad P(\text{pos}\wedge\text{clean}) \:=\:0.00995 \;\;{\color{blue}[2]}$

From [1] and [2]: .$\displaystyle P(\text{pos}) \:=\:0.00475 + 0.0095 \:=\:0.0147\;\;{\color{blue}[3]}$

"What is the probability that an athlete who tests positive did use the drug?"

. . We want: .$\displaystyle P(\text{use}|\text{pos}) \:=\:\frac{P(\text{use}\wedge\text{pos})}{P(\text{ pos})}$

From [1] and [3]: .$\displaystyle P(\text{use}|\text{pos}) \:=\:\frac{0.00475}{0.0147} \;\approx\;0.323 \;=\;32.3\%$

4. Whether that's right or not, that deserves a thanks

Although only 32% of people who test positive actually use?? That seems a little unrealisic to me...

5. Originally Posted by chella182
Whether that's right or not, that deserves a thanks

Although only 32% of people who test positive actually use?? That seems a little unrealisic to me...