I always believed that problems on conditional probability could be solved with common logic without using Bayes' theorem (because I cannot understand Bayes' theorem intuitively and I didn't bother because I knew another way).
But this problem gives me varying answers with general logic and Bayes' Theorem.
A test for a sickness is 99% reliable. That is, 99% of the healthy people test negative, and 99% of sick people test positive.
Given that 1% of the entire population is actually sick, what are the chances that you are actually sick if you test positive?
Here is my solution without using Bayes' theorem.
$$P(being\ sick\ |\ test\ positive) = \frac{Number\ of\ people\ who\ are\ actually\ sick}{Number\ of\ people\ who\ test\ positive} $$
$$So,P(being\ sick\ |\ test\ positive)= \frac{1\%\ of\ \ the\ total\ population}{1\%\ of\ healthy\ people+99\%\ of\ sick\ people} $$
$$Let\ the\ total\ population\ be\ x. So,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $$
$$P=\frac{0.01x}{0.01*0.99x+0.99*0.01x}$$
$$P=\frac{1}{0.99+0.99}=\frac{1}{1.98}$$
But, when you use Bayes' theorem, you get P=1/2. Why is the denominator off by 0.02?
Either I'm wrong (but I really can't see why) or there is some kind of an approximation in Bayes' theorem (highly impossible). Can someone please tell me what are the other factors that Bayes' theorem considers but my solution doesn't? I would also be extremely grateful if someone could explain the theorem can be understood intuitively.
Best Answer
Your numerator should be those people that get sick AND test positive, or $0.99\times 0.01x$.
In some sense, your universe is those people who test positive, which is $0.01x$. That includes healthy people with false positives, and sick people with true positives. In this problem those two populations are of equal size, hence $\frac{1}{2}$.