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JEE Advanced 2021 Paper 1

Section: Mathematics

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Question : 41 of 57

Marks: +1, -0

Consider three sets

E_{1}=\\{1,2,3\\}, F_{1}=\\{1,3,4\\}

G_{1}=\\{2,3,4,5\\}

. Two elements are chosen at random, without replacement, from the set

E_{1}

S_{1}

denote the set of these chosen elements.

E_{2}=E_{1}-S_{1}

F_{2}=F_{1} \cup S_{1}

. Now two elements are chosen at random, without replacement, from the set

F_{2}

S_{2}

denote the set of these chosen elements.

G_{2}=G_{1} \cup S_{2}

. Finally, two elements are chosen at random, without replacement, from the set

G_{2}

S_{3}

denote the set of these chosen elements.

E_{3}=E_{2} \cup S_{3}

E_{1}=E_{3}

p

be the conditional probability of the event

S_{1}=\\{1,2\\} .

Then the value of

p

[JEE Adv 2021 P1]

$\frac{1}{5}$
$\frac{3}{5}$
$\frac{1}{2}$
$\frac{2}{5}$

Solution:

P=\frac{P(S_{1} \cap (E_{1}=E_{3}))}{P(E_{1}=E_{3})}=\frac{P(A_{1,2})}{P(A)}

P(A)=P(A_{1,2})

+P(A_{1,3})

+P(A_{2,3})

\underset{\text{If 1,2 chosen at start}}{\uparrow}

\underset{}{\underset{\text{If 1,2 chosen at start}}{\uparrow}}

\uparrow_{\text{If 1,2 chosen at start}}

P(A_{1,2})=\frac{1}{3} \times \frac{1 \times {}^3 C_1}{{}^4 C_2}_{\text{1 is definitely chosen from}\\; F_{2}} \times \frac{1}{{}^5 C_2}_{\text{1,2 chose from}\\; G_{2}}

=\frac{1}{3} \times \frac{1}{2} \times \frac{1}{10}=\frac{1}{60}

P(A_{1,3})=\frac{1}{3} \times \frac{1 \times {}^2 C_1}{{}^3 C_2}_{\text{1 is definitely chosen from}\\; F_{2}} \times \frac{1}{{}^5 C_2}_{\text{1,2 chosen from } G_{2}}

=\frac{1}{3} \times \frac{2}{3} \times \frac{1}{10}=\frac{1}{45}

P(A_{2,3})=\frac{1}{3} \times \left[ \frac{{}^3 C_2 \times 1}{{}^4 C_2}_{\text{If 1 is not chosen from}\\; F_{2}} \times \frac{1}{{}^4 C_2} + \frac{1 \times {}^3 C_1}{4 C_{2}}_{\text{If 1 is not chosen from}\\; F_{2}} \times \frac{1}{{}^4 C_2} \right]

=\frac{1}{3}\left(\frac{1}{12}+\frac{1}{20}\right)=\frac{2}{45}

P(A)=\frac{1}{60}+\frac{1}{45}+\frac{2}{45}=\frac{1}{12}

\frac{P(A_{1,2})}{P(A)}=\frac{1}{5}

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