Question

A slip of paper is given to a person A who marks it either with a plus sign or a minus sign. The probability of his writing a plus sign is $\frac{1}{3}$. A passes the slip to B, who may either leave it alone or change the sign before passing it to C. Next C passes the slip to D after perhaps changing the sign. Finally D passes it to a referee after perhaps changing the sign. B, C, D each change the sign with probability $\frac{2}{3}$.
If the referee observes a plus sign on the slip then the probability that A originally wrote a plus sign is

• A. $\frac{14}{27}$
• B. $\frac{13}{41}$
• C. $\frac{13}{27}$
• D. $\frac{27}{14}$

Answer 1

The correct option is B $\frac{13}{41}$

Let $E_1$ = Event that A wrote a plus sign.
$E_2$ = Event that A wrote a minus sign.
E = Event that the referce observes a plus sign.
$GivenP\left(E_1\right)=\frac{1}{3}\Rightarrow P\left(E_2\right)=\frac{2}{3}$
$P\left(\frac{E}{E_1}\right)$ = Probability that none of B, C, D change sing + Probability that exactly two of B, C, D change sign.
$=\frac{1}{27}+3(\frac{1}{3}\times \frac{2}{3}\times \frac{2}{3})=\frac{13}{27}$
$P\left(\frac{E}{E_2}\right)$= Probability that all of B, C, D change the sign + Probability that exactly one of them changes the sign.
$=\frac{8}{27}+3\times (\frac{2}{3}\times \frac{1}{3}\times \frac{1}{3})=\frac{14}{27}$
$\therefore P\left(\frac{E_1}{E}\right)=\frac{13}{41}$ Using Baye’s theorem.