JEE Advanced 2021 Paper 1 · Q03 · Probability

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

SolveKar AI · Accepted Answer

For $E_1=E_3$ we need $S_3$ to supply the elements removed in $S_1$ via $E_2$. Three disjoint cases for $S_1$: (i) $\{1,2\}$: $P_1=\dfrac{1}{60}$ (the $1$ must persist in $F_2$ and $2$ must come back via $S_3$). (ii) $\{2,3\}$: $P_2=\dfrac{2}{45}$. (iii) $\{1,3\}$: $P_3=\dfrac{1}{45}$ (since $1,3$ are already in $F_1$, $F_2$ stays $\{1,3,4\}$). Required conditional probability $=\dfrac{P_1}{P_1+P_2+P_3}=\dfrac{1/60}{1/60+2/45+1/45}=\dfrac{1}{5}$. Hence (A).

JEE Advanced 2021 Paper 1 · Q03 · Probability

Solution