JEE Advanced 2021 · Paper 1 · Mathematics Q3

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MathematicsSingle correct+3 / -1Question 41 of 57

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

\frac15

\frac35

\frac12

\frac25

Answer and worked solution

Official answer: A

When does $E_3=E_1$ ? $E_2$ has one element and $S_3$ has two, so $E_3=E_1$ exactly when $S_3=S_1$ .

Case by case.

$S_1=\{1,2\}$ : $F_2=\{1,2,3,4\}$ . For $1\in G_2$ , $S_2$ must contain 1 ( $\frac36$ ); then $G_2$ has 5 elements and $P(S_3=\{1,2\})=\frac1{10}$ . Contribution: $\frac13\cdot\frac12\cdot\frac1{10}=\frac1{60}$ .
$S_1=\{1,3\}$ : $F_2=\{1,3,4\}$ , and $S_2$ must contain 1 ( $\frac23$ ). Contribution: $\frac13\cdot\frac23\cdot\frac1{10}=\frac1{45}$ .
$S_1=\{2,3\}$ : $F_2=\{1,2,3,4\}$ . If $S_2\subset G_1$ ( $\frac12$ ), $G_2$ has 4 elements and the chance is $\frac16$ ; otherwise it has 5 elements and the chance is $\frac1{10}$ . Contribution: $\frac13\left(\frac1{12}+\frac1{20}\right)=\frac2{45}$ .

Conditional probability. In total $P(E_3=E_1)=\frac1{60}+\frac3{45}=\frac1{12}$ , so
$p=\frac{1/60}{1/12}=\frac15.$
(A)

Topics:Conditional probability Bayes' theorem

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