Question

Consider four independent trials in which an event $A$ occurs with probability $\frac{1}{3}$. The event $B$ will occur with probability $1$ if the event $A$ occurs at least twice, it can not occur if the event $A$ does not occur and it occurs with a probability $\frac{1}{2}$ if the event $A$ occurs once. If the probability $p$ of the occurrence of event $B$ can be expressed as $\frac{m}{n},$ where $m,n\in N,$ then the least value of $m+n$ is

Answer 1

$n=4;P\left(A\right)=\frac{1}{3};P\left(B\right)=p$
Let $E_0=$ event $A$ does not occur
$E_1=$ event $A$ occurs once
$E_2=$ event $A$ occurs at least twice

$P\left(B|E_0\right)=0$
$P\left(B|E_1\right)=\frac{1}{2}$
$P\left(B|E_2\right)=1$
By total probability theorem,
$P\left(B\right)=P(E_0\cap B)+P(E_1\cap B)+P(E_2\cap B)$
$=P\left(E_0\right)P\left(B|E_0\right)+P\left(E_1\right)P\left(B|E_1\right)++P\left(E_2\right)P\left(B|E_2\right)$
$=0+{}^4{C}_1\cdot \frac{1}{3}\cdot {\left(\frac{2}{3}\right)}^3\cdot \frac{1}{2}+({}^4{C}_2\cdot {\left(\frac{1}{3}\right)}^2\cdot {\left(\frac{2}{3}\right)}^2+{}^4{C}_3\cdot {\left(\frac{1}{3}\right)}^3\cdot \frac{2}{3}+{}^4{C}_4\cdot {\left(\frac{1}{3}\right)}^4)\cdot 1$
$=\frac{16+24+8+1}{81}=\frac{49}{81}=\frac{m}{n}$
$\Rightarrow (m+n{)}_{min}=130$