Mathematics · Probability

JEE Advanced 2021 — Paper 1 — Question 21

Consider three sets $\mathrm{E}_{1}=\{1,2,3\}, \mathrm{F}_{1}=\{1,3,4\}$ and $\mathrm{G}_{1}=\{2,3,4,5\}$ .

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 $\mathrm{S}_{3}$ 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

Answer: C

E_1=\{1,2,3\},\qquad |E_1|=3 \Rightarrow |S_1|=2,\ \text{total\; choices\;}=3

\text{Case\; } S_1=\{1,2\}:\ E_2=\{3\},\ F_2=\{1,2,3,4\}

|F_2|=4 \Rightarrow \binom{4}{2}=6,\ \text{favourable\; outcomes\; for\; }E_3=E_1=3

P(E_3=E_1\mid S_1=\{1,2\})=\frac{3}{6}=\frac12

\text{For\; } S_1=\{1,3\}\ \text{or}\ \{2,3\}: \quad P(E_3=E_1\mid S_1)=\frac14

P(E_3=E_1)=\frac13\!\left(\frac12+\frac14+\frac14\right)=\frac13

p=P(S_1=\{1,2\}\mid E_3=E_1) =\frac{\frac13\cdot\frac12}{\frac13} =\boxed{\frac12}

Answer key and solution verified before publishing.

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