Question

Consider three sets $E_1=\{1,2,3\},F_1=\{1,3,4\}$ and $G_1=\{2,3,4,5\}$. Two elements are chosen at random, without replacement, from the set $E_1$ and let $S_1$ denote the set of these chose elements. Let $E_2=E_1-S_1$ and $F_2=F_1\cup S_1$. Now two elements are chosen at random, without replacement, from the set $F_2$ and let $S_2$ denote the set of these chosen elements.
Let $G_2=F_1\cup S_2$. Finally, two elements are chosen at random, without replacement, from the set $G_2$ and let $S_3$ denote the set of these chosen elements.
Let $E_3=E_2\cup S_3$. Given that $E_1=E_3$, let $p$ be the conditional probability of the event $S_1=\{1,2\}$. Then the value of $p$ is

• A. $\frac{1}{5}$
• B. $\frac{3}{5}$
• C. $\frac{1}{2}$
• D. $\frac{2}{5}$

Answer 1

Answer: (A)

	$S_1$	$F_2=F_1\cup S_1$	$S_2$	$G_1\cup S_2=G_2$	$S_3$
(i)	$\{1,2\}$	$\{1,2,3,4\}$	$\{1,x\}$	$\{1,2,3,4,5\}$	$\{1,2\}$
(ii)	$\{2,3\}$	$\{1,2,3,4\}$	$\{1,x\}$	$\{1,2,3,4,5\}$	$\{2,3\}$
			$\{x,y\}$ (where $x$ and $y$	or $\{2,3,4,5\}$
(iii)	$\{1,3\}$	$\{1,3,4\}$	$\{1,x\}$ are other than 1$)$	$\{1,2,3,4,5\}$	$\{1,3\}$

(i) $P_1=\frac{{}^2{c}_2}{{}^3{c}_2}\cdot \frac{{}^3{c}_1}{{}^4{c}_2}\cdot \frac{{}^2{c}_1}{{}^5{c}_2}=\frac{1}{60}$
(ii) $P_2=\frac{1}{3}\left(\frac{{}^3{c}_1}{{}^4{c}_2}\times \frac{{}^2{c}_2}{{}^5{c}_2}+\frac{{}^3{c}_2}{{}^4{c}_2}\times \frac{{}^2{c}_2}{{}^4{c}_2}\right)=\frac{2}{45}$
(iii) $P_3=\frac{1}{{}^3{c}_1}\times \frac{{}^2{c}_1}{{}^3{c}_2}\times \frac{{}^2{c}_2}{{}^5{c}_2}=\frac{1}{45}$
Conditional probability $=\frac{P_1}{P_1+P_2+P_3}=\frac{1}{5}$