I'm not sure what I don't get. The chances of a man having prostate cancer is 1/14, and the chance for them to not have high PSA is 7% so I multiplied them and got 0.5%. What am I missing?
#Probability question (Answer is 0.7%)
rugged talon
latent oysterBOT
I'm not sure what I don't get. The chances of a man having prostate cancer is 1/...
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rugged talon
Probability question (Answer is 0.7%)
bright peak
I'm not sure what I don't get. The chances of a man having prostate cancer is 1/...
Have you heard of Bayes's Theorem?
rugged talon
Nope.
bright peak
Okay, let's track what probabilities we're actually given, and what precisely they're probabilities of. So what's the first figure we're given?
rugged talon
Well, we know that at least 1 out of every 14 man has prostate cancer above the age of 50.
The man we're given is over 50 so we don't have to involve any chance regarding his age.
bright peak
Right, so let's just say P(cancer) = 1/14.
rugged talon
Next, we know that there's a 7% chance that a person with prostate cancer will not have a high PSA.
This must be the case as well.
bright peak
P(~PSA|cancer) = 0.07
rugged talon
?
bright peak
What's wrong?
bright peak
~ = "not", | = "given"
rugged talon
Ahh
So that means, probability of not having PSA among people who absolutely have cancer?
bright peak
Yeah.
rugged talon
Which would be, 7% x 1/14 yeah?
bright peak
...no.
bright peak
There's another number given.
bright peak
It's good practice to always assume all information given in the question is relevant.
rugged talon
Ironic, because in the other sections of the test the opposite is advised (and generally holds true)
But I'll keep it in mind, it will not be disregarded until I prove it is worthless.
What do we do with it, how does it relate?
rugged talon
Fair enough.
- 1/14 chance for a man over 50 to have prostate cancer
- 7% chance for a man with cancer to not have high psa
- 75% chance for a man with high psa to not have cancer
Our man has normal PSA, and we want to know how likely he is to have prostate cancer.
bright peak
P(cancer) = 1/14
P(~PSA|cancer) = 0.07
P(~cancer|PSA) = 0.75```What we want is P(cancer|~PSA)
bright peak
So here's a formula for conditional probability: P(A|B) = P(A&B)/P(B)
rugged talon
So the chance of cancer, given that he doesn't have psa, is = P(has cancer AND low psa) / P(low psa)?
bright peak
Right.
rugged talon
P(cancer and psa) is, then 7% x 1/14 ?
bright peak
Not necessarily.
That only works if they're independent, which the fact that this is used as a test implies they're not, otherwise it'd be kind of useless.
bright peak
So here's where we might invoke Bayes's Theorem. We still have a few problems if we do, but it might make more sense. I'll actually derive it for you.
That should help you remember it. It helps me.
So to start with, if $P(A|B) = \frac{P(A&B)}{P(B)}$, then logically $P(A&B) = P(A|B)P(B)$, right?
rugged talon
Yeah, that checks out
sacred sirenBOT
techieliterate
bright peak
But A and B are just placeholders, so as long as we replace all the As with the same thing and all the Bs with the same thing, the formula holds. Let's replace all the As with Bs and all the Bs with As: $P(B&A) = P(B|A)P(A)$. But of course, $P(B&A) = P(A&B)$.
sacred sirenBOT
techieliterate
rugged talon
Also checks out
bright peak
Therefore, $P(A|B) = \frac{P(B|A)P(A)}{P(B)}$.
sacred sirenBOT
techieliterate
bright peak
Now, you can check it against your intuition that $P(B) = P(A&B) + P(~A&B)$, right?
sacred sirenBOT
techieliterate
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rugged talon
Nope, brainfarted there
bright peak
Okay, P(A&B) is the probability that A happens and B happens, right?
rugged talon
Sorry, I'm very out of practice with math and I've been trying my best, but relearning is hard
rugged talon
Okay, `P(A&B)` is the probability that A happens and B happens, right?
Yeah!
Both happen
bright peak
P(~A&B) is correspondingly the probability that A does not happen and B happens.
Whenever B happens, A either happens or doesn't happen, and it can't happen and not happen, therefore every instance of B is accounted for in exactly one of P(A&B) or P(~A&B).
This diagram might help:
bright peak
Both happen
Are you following?
bright peak
Well, we just got done making a substitution for $P(A&B)$, so we can use that here: $P(B) = P(A&B)+P(~A&B) = P(B|A)P(A) + P(B|~A)P(~A)$
sacred sirenBOT
techieliterate
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bright peak
Thus, Bayes's Theorem:
$P(A|B) = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|~A)P(~A)}$
sacred sirenBOT
techieliterate
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bright peak
$P(A|B) = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|~A)P(~A)}$
sacred sirenBOT
techieliterate
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bright peak
There!
rugged talon
And with this theorem, we do...
bright peak
Well, we want P(cancer|~PSA), so... plugging in...
rugged talon
$\frac{P(~PSA|Cancer)P(Cancer)}{P(~PSA|Cancer)P(Cancer) + P(~PSA|~Cancer)P(~Cancer)}$
sacred sirenBOT
fuelweaver
bright peak
Forgot the backslashes on the "not"s.
rugged talon
Hmm, didn't read 'em
bright peak
$\frac{P(~{PSA}|Cancer)P(Cancer)}{P(~{PSA}|Cancer)P(Cancer)+P(~{PSA}|~{Cancer})P(~{Cancer})}$
sacred sirenBOT
techieliterate
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bright peak
...dammit.
rugged talon
P(cancer) = 1/14
P(~PSA|cancer) = 0.07
P(~cancer|PSA) = 0.75
(0.07 x 1/14) / (0.07 x 1/14) + whuh
There's gotta be a simpler way to do this than the theorem
bright peak
Not really. Though we don't exactly have everything we need for the theorem yet.
bright peak
Okay, look. There's a simpler form of Bayes's Theorem, but it's in odds, not probabilities, so we'd have to convert back once we were done.
rugged talon
Not really. Though we don't exactly have everything we need for the theorem yet.
Hmm, how do we get the right side of the bottom?
bright peak
Hmm, how do we get the right side of the bottom?
That's exactly the problem. I've got it now, but it's taken me a good deal of finagling.
Or rather, I've obviated my need for it by finding something else.
rugged talon
Maybe we should try to figure out the "simple" solution I linked?
bright peak
Actually, that's the wrong solution according to my math.
rugged talon
Gah. Nothing makes sense anymore.
bright peak
Look, we know that P(cancer|~PSA) = P(cancer & ~PSA)/P(~PSA), right?
rugged talon
Yeah!
bright peak
So far, what we have is P(cancer), P(~PSA|cancer), and P(~cancer|PSA), right?
rugged talon
yeah.
bright peak
So we just need to use what we have to get what we want.
We know P(~PSA|cancer) * P(cancer) = P(cancer & ~PSA), so we have that down.
rugged talon
What we're lacking is P(~PSA)
bright peak
Right.
rugged talon
Which is just 1 - P(PSA) yeah?
bright peak
...yes, but we don't have P(PSA) either, so that doesn't strictly help.
rugged talon
So we're trying to get P(PSA)
bright peak
Right.
rugged talon
7% of 1/14 people have low PSA, 93% OF 1/14 People have high psa.
the 93% of 1/14 makes up the 25% of men with high PSA
bright peak
...no?
rugged talon
Why not?
bright peak
...because the 93% of the 1/14 have cancer. The 25% don't.
rugged talon
The 75% don't
bright peak
...oh.
Right.
Hang on, I think I see where you're going. Let me just write it down in notation to check it.
rugged talon
So the amount of people with high PSA would be 4/14 * 93%
around 26%
but something tells me we're not supposed to get numbers like these on a no-calculator test
bright peak
So the amount of people with high PSA would be 4/14 * 93%
It's actually 2/7 * 93%.
rugged talon
Literally the exact same thing
rugged talon
So what do we do now?
bright peak
Well, we have P(PSA), so we subtract it from 1.
bright peak
P(cancer|~PSA) = P(cancer & ~PSA)/P(~PSA) = (0.07/14)/(5.14/7)
bright peak
...well, no?
rugged talon
from that, we get 0.007 or 0.7%
bright peak
Well. We round to it. The fact that we don't get an exact answer makes the official explanation seem even hackier.
rugged talon
This was the first exam with like 5 participants so I guess no one objected to how wacky this was for a logical reasoning question
I will probably never see it again
but it's good to learn just in case
bright peak
You actually taught me a few things, namely that trick where P(A) = P(A|B)*P(B)/P(B|A). That was cool.