Question

There is a raffle with $10$ tickets and two prizes of value $5$ pounds and $1$ pounds respectively. A holds one ticket and is informed by B that he has won the $5$ pounds prize, while C asserts that he has won the $1$ pounds prize: what is A's expectation, if the credibility of B is denoted by $\frac{2}{3}$, and that of C by $\frac{3}{4}$?

Answer 1

10 Tickets of two prizes value 5 pounds and 1 pound.

B's credibility $=\frac{2}{3}$

C's credibility $=\frac{3}{4}$

$P\left(A\right)=P\left(E_1\right)\left(\frac{A}{E_1}\right)+P\left(E_2\right)\left(\frac{A}{E_2}\right)+P\left(E_3\right)\left(\frac{A}{E_3}\right)+.....P\left(E_n\right)\left(\frac{A}{E_n}\right)$

Case(1): If B and C are telling truth,then

chance of winning A=0

So, $P\left(E_1\right)\left(\frac{A}{E_1}\right)=\frac{2}{3}\times \frac{3}{4}\times 0$

Case(2): If B is telling truth and C is not

chance of winning A= $\frac{1}{9}$

So, $P\left(E_2\right)\left(\frac{A}{E_2}\right)=\frac{2}{3}\times \frac{1}{4}\times \frac{1}{9}$

Case(3): If C is telling truth and B is not

So, $P\left(E_3\right)\left(\frac{A}{E_3}\right)=\frac{1}{3}\times \frac{3}{4}\times \frac{1}{9}$

Case(4): Both are lying then chance of A=$\frac{2}{10}$

chance of winning A= $\frac{1}{9}$

So, $P\left(E_2\right)\left(\frac{A}{E_2}\right)=\frac{2}{3}\times \frac{1}{4}\times \frac{2}{10}$

$P\left(A\right)=(\frac{2}{3}\times \frac{3}{4}\times 0)+(\frac{2}{3}\times \frac{1}{4}\times \frac{1}{9})+(\frac{1}{3}\times \frac{3}{4}\times \frac{1}{9})+(\frac{2}{3}\times \frac{1}{4}\times \frac{2}{10})=\frac{17}{270}$