An urn contains n balls each identical in size and texture but with different colors. One of the balls is white. Two independent observers, each having a probability of 0.1 of telling the truth, assert that a ball drawn at random from the urn is white. Prove The probability that the ball is in fact white is (n-1)/(n+80).
I have proceeded as this: E1: Event that both observers are telling the truth E2: both are lying
E: A white draw is reported
p(E/E1) =1/n P(E/E2)=n-1/n P(e1)=1/100 P(E2)=811/100. Hence probablity that a white ball was drwan given that a white draw was reported was p(E1/E)= P(e/E1)*P(E1)/P(e/e1)*p(e1)+P(e/e2)*p(e2).. but i am not arriving at the answer..where am i going wrong?
The probability both are lying may be $\frac{81}{100}$ but they need not report "white" when lying as there are $n-1$ possible colours they could each say when lying.
Indeed, if the actual colour is not-white and they are both lying, the conditional probability that a particular one of them says "white" is $\frac{1}{n-1}$ and so that both say "white" is $\left(\frac{1}{n-1}\right)^2$. You need to include the conditional probability that they both actually say "white" in your calculation, making it
$$ \dfrac {\frac{1}{n}\times \frac{1}{100} \times 1 }{\frac{1}{n}\times \frac{1}{100}\times 1 \;+\; \frac{n-1}{n}\times \frac{81}{100}\times \left(\frac{1}{n-1}\right)^2 } $$
and this will give the expected answer.