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 chosen 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 $G_2=G_1\cup S_2.$ 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

The correct option is A $\frac{1}{5}$

We will follow the tree diagram.


$P(E_1=E_3)=\frac{1}{3}(\frac{1}{2}\times \frac{1}{10}+\frac{1}{2}\times 0+\frac{1}{2}\times \frac{1}{10}+\frac{1}{2}\times \frac{1}{6}+\frac{2}{3}\times \frac{1}{10}+\frac{1}{3}\times 0]=\frac{1}{3}\left[\frac{1}{4}\right)$
Required probability $=\frac{\frac{1}{3}(\frac{1}{2}\times \frac{1}{10})}{\frac{1}{3}\times \frac{1}{4}}=\frac{1}{5}$