Question

A slip of paper is given to $A$, who marks it with either a plus or a minus sign; the probability of his writing a plus is $\frac{1}{3}$. He then passes the slip to $B$, who may either leave it or change the sign before passing it on to $C$. Next $C$ passes the slip to $D$ after perhaps changing the sign; finally $D$ passes it to an honest judge after perhaps changing the sign. The judge sees a plus sign on the slip. It is known that $B,C$ and $D$ each change the sign with probability $\frac{2}{3}$. Then probability that $A$ originally wrote a plus is $\frac{a}{b}$ (where $a,b$ are coprime numbers), then the value of $a+b$ is

Answer 1

$x\to$ A left by plus sign

$y\to$ final result was a plus sign

$P\left(\frac{x}{y}\right)=\frac{P(x\cap y)}{P\left(y\right)}$

for final plus sign, the sign must have changed even number of times in the process. So, there can be $0,2,\text{or}4$ times sign can be changed.

0 times $\to {\left(\frac{1}{3}\right)}^4$

4 times $\to {\left(\frac{2}{3}\right)}^4$

2 times $\to {}^4{C}_2{\left(\frac{2}{3}\right)}^2{\left(\frac{1}{3}\right)}^2$

So $p\left(y\right)=\frac{1}{81}+\frac{16}{81}+\frac{24}{81}=\frac{41}{81}$

For $p(x\cap y)$, the process must start with positive sign and number of changes must be even. As only $B,C,D$ can change the sign, only $0$ or $2$ times sign can be changed

Probability of $0$ changes is ${\left(\frac{1}{3}\right)}^4$

and $2$ changes is ${}^3{C}_2{\left(\frac{2}{3}\right)}^2{\left(\frac{1}{2}\right)}^2=\frac{12}{81}$

$p(x\cap y)=\frac{1}{81}+\frac{12}{81}=\frac{13}{81}$

$p\left(\frac{x}{y}\right)=\frac{13}{41}$