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 $=\frac{P_{1}}{P_{1}+P_{2}+P_{3}}=\frac{1}{5}$