Conditional probability, Likelihood The probability of an issue being assigned to tester $A$ is $0.6$, and to tester $B$ is 0.4. The probabilty of tester A finding an error is $0.94$ and $0.98$ for tester $B$. If an error was found what is the probability that the issue was assigned to tester $A$.

$E = $an error was found.

$A = $issue assigned to tester A.

$A^\complement = $issue assigned to tester B.

$P(A|E)=\frac{P(E|A)P(A)}{P(E|A)P(A)+P(E|A^\complement)P(A^\complement)}$

$P(A|E)=\frac{\frac{94}{100}\frac{6}{10}}{\frac{94}{100}\frac{6}{10}+\frac{98}{100}\frac{4}{10}}=\frac{\frac{564}{1000}}{\frac{956}{1000}}=\frac{564}{956}=\frac{141}{239}$