Mathematics - Probability Question with Solution | TestHub

We need to find $p$, the conditional probability of the event $S_1=\{1,2\}$, given $E_1=E_3$.

So, $p=\frac{P\left(S_1\cap \left(E_1=E_3\right)\right)}{P\left(E_1=E_3\right)}=\frac{P\left(B_{1,2}\right)}{P\left(B\right)}$

$P\left(B\right)=P\left(B_{1,2}\right)+P\left(B_{1,3}\right)+P\left(B_{2,3}\right)$

$P\left(B_{1,2}\right)\to$ If $\{1,2\}$ chosen in the beginning 

$P\left(B_{1,3}\right)\to$If $\{1,3\}$ chosen in the beginning

$P\left(B_{2,3}\right)\to$If $\{2,3\}$ chosen in the beginning

(i) If $\{1,2\}$ is chosen in the begining, then $F_2=\{1,2,3,4\}$ and $S_2$ must contain $\left\{1\right\}$ and $S_3=\left\{1,2\right\}$ from $G_2=\{1,2,3,4,5\}$

So, $P\left(B_{1,2}\right)=\frac{1}{3}\times \frac{1\times {}^{3}C_1}{{}^{4}C_2}\times \frac{1}{{}^{5}C_2}=\frac{1}{3}\times \frac{1}{2}\times \frac{1}{10}$

(ii) If $\{1,3\}$ is chosen in the begining, then $F_2=\{1,3,4\}$ and $S_2$ must contain $\left\{1\right\}$ and $S_3=\left\{1,3\right\}$ from $G_2=$$\{1,2,3,4,5\}$

So, $P\left(B_{1,3}\right)=\frac{1}{3}\times \frac{1\times {}^{2}C_1}{{}^{3}C_2}\times \frac{1}{{}^{5}C_2}=\frac{1}{3}\times \frac{2}{3}\times \frac{1}{10}$

(iii) If $\{2,3\}$ is chosen in the begining, then $F_2=\{1,2,3,4\}$, then two cases are possible

1. $S_2$ may contain $\left\{1\right\}$ and getting $S_3=(2,3)$ from $G_2=\{1,2,3,4,5)$

2.  $S_2$ does not contain $\left\{1\right\}$ and getting $(2,3)$ from $G_2=\{2,3,4,5]$

So, $P\left(B_{2,3}\right)=\frac{1}{3}\times \left[\frac{{}^3\mathrm{C}_2\times 1}{{}^4\mathrm{C}_2}\times \frac{1}{{}^4\mathrm{C}_2}+\frac{1\times {}^3\mathrm{C}_1}{{}^4\mathrm{C}_2}\times \frac{1}{{}^5\mathrm{C}_2}\right]=\frac{1}{3}\times \left[\frac{1}{2}\times \frac{1}{6}+\frac{1}{2}\times \frac{1}{10}\right]$

Now, $p=\frac{P\left(B_{1,2}\right)}{P\left(B_{1,2}\right)+P\left(B_{1,3}\right)+P\left(B_{2,3}\right)}$

$=\frac{\frac{1}{3}\times \frac{1}{2}\times \frac{1}{10}}{\frac{1}{3}\times \frac{1}{2}\times \frac{1}{10}+\frac{1}{3}\times \frac{2}{3}\times \frac{1}{10}+\frac{1}{3}\times \left[\frac{1}{2}\times \frac{1}{6}+\frac{1}{2}\times \frac{1}{10}\right]}=\frac{1}{5}$