Apples, Oranges & Bayes Theorm

Probability of having a true positive given that you tested positive

$P(+|D) = 99\%$

$P(+|\bar D) = 2\%$

$P(D)=1\%$

$P(D|+) = \frac{P(+|D)*P(D)}{P(+)} = \frac{P(+|D)P(D)}{P(+|D)P(D) \; + \; P(+|\bar D)P(\bar D)} =$ $\frac{0.990.01}{0.990.01+0.020.99}=\frac{0.01}{0.01+0.02}=\frac{1}{3}$

Quiz 1

Probability vs Density