Cancer Test
1% of people have a rare cancer, and there is a test for this cancer which is "90% accurate"; that is:
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If you have the cancer there is a 90% chance the test will be positive.
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If you don't have the cancer there is a 90% chance the test will be negative.
If you take the test and test positive, what is the approximate probability that you have the cancer?
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4 solutions
Well the options should be proper... otherwise there remains an element of doubt even after doing it correctly
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I'm glad you experienced doubt fyi many doctors get this wrong :/ imagine being the patient in that situation! Doubt is a good thing and cuases you to go back and make sure. You can also rule out the wrong answers to be sure for example 1% would have to be like 9/900
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Fair enough that it did make me go back over and check my answer a couple of times, but we don't need to be given FALSE answers to teach us to double check things?
The solution is wrong. It clearly states that there is a 90% chance you have the cancer if you test positive. Therefore, the answer should be 90%, unless you mean "the probability that you DON'T have the cancer", in which case 10% would be the correct answer.
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Actually, it's the other way round: it says that if you have cancer there's a 90% chance you'll test positive.
Well, it's not wrong.
He could write the entire solution using
P(C|T) = P(C & T)/P(T) ... and he didn't.
Which would lead to the same result.
The results for the test are:
P(T|C) = 90% P(¬T|¬C) = 90%
Where T means: did positive C means: does have cancer
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It is actually wrong. Puzzle constructor confused him/herself. Sometimes there is an upper logic where formula is useless
Use Bayes theorem
P ( C a n c e r ∣ P o s i t i v e ) = P ( P o s i t i v e ) P ( P o s i t i v e ∣ C a n c e r ) ⋅ P ( C a n c e r ) = Let's call Positive = P, Cancer = C, No Cancer = ¬C P ( P ∣ C ) ⋅ P ( C ) + P ( P ∣ ¬ C ) ⋅ P ( ¬ C ) 1 0 0 1 ⋅ 1 0 9 = = 1 0 0 0 9 + ( 1 0 0 9 9 ⋅ 1 0 1 ) 1 0 0 0 9 = 1 0 0 0 9 + 1 0 0 0 9 9 1 0 0 0 9 = 1 0 8 9 = 1 2 1 ≈ 1 0 %
Use bayers theoram P(affected/+ve result)=[90/100× 1/100]÷[ (90/100 ×1/100)+ (10/100×99/100)]
The answers should have an approximation sign, to reinforce that the question asks about the "approximate probability that you have the cancer."
Imagine 1000 people.
1% means 10 people have cancer.
Of those 10, 90% or 9 will test positive.
Of the 990 people that don't have cancer 10% or 99 will test false positive.
108 will test positive, but only 9 will have cancer.
So its 9/108 chance or a little less than 10%