Consider three sets E _ 1 = 1, 2, 3 , F _ 1 = 1, 3, 4 and G _ 1 =...

We need to find $p$, the conditional probability of the event ${S}_{1} = { 1, 2 }$, given ${E}_{1} = {E}_{3}$.

So, $p = \dfrac{P\left({S}_{1} \cap \left({E}_{1} = {E}_{3}\right)\right)}{P\left({E}_{1} = {E}_{3}\right)} = \dfrac{P\left({B}_{1, 2}\right)}{P ( B )}$

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

$P\left({B}_{1, 2}\right) \to$ 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 }$</p><p>So,$P\left({B}_{1, 2}\right) = \dfrac{1}{3} \times \dfrac{1 \times C1 3}{C2 4} \times \dfrac{1}{C2 5} = \dfrac{1}{3} \times \dfrac{1}{2} \times \dfrac{1}{10}$</p><p><strong>(ii)</strong> If${ 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, $P\left({B}_{1, 3}\right) = \dfrac{1}{3} \times \dfrac{1 \times C1 2}{C2 3} \times \dfrac{1}{C2 5} = \dfrac{1}{3} \times \dfrac{2}{3} \times \dfrac{1}{10}$

(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 )$</p><p>2.${S}_{2}$does not contain \left\{1\right\}$ and getting $( 2, 3 )$ from {G}_{2} = { 2, 3, 4, 5 ]

So, $P\left({B}_{2, 3}\right) = \dfrac{1}{3} \times \left[\dfrac{C2 3 \times 1}{C2 4} \times \dfrac{1}{C2 4} + \dfrac{1 \times C1 3}{C2 4} \times \dfrac{1}{C2 5}\right] = \dfrac{1}{3} \times \left[\dfrac{1}{2} \times \dfrac{1}{6} + \dfrac{1}{2} \times \dfrac{1}{10}\right]$

Now, $p = \dfrac{P\left({B}_{1, 2}\right)}{P\left({B}_{1, 2}\right) + P\left({B}_{1, 3}\right) + P\left({B}_{2, 3}\right)}$

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