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QJEEAdv21Ma99e4: JEE Advanced 2021 Mathematics - Class 12 JEEAdv21Ma99e4 Solution: Consider three sets E 1 = {1, 2, 3}, F 1 = {1, 3, 4} and G 1 =…

Consider three sets E1 = {1, 2, 3}, F1 = {1, 3, 4} and G1 = {2, 3, 4, 5}. Two elements are chosen at random, without replacement, from the set E1, and let S1 denote the set of these chosen elements. Let E2 = E1 −- S1 and F2 = F1 ∪\cup S1. Now two elements are chosen at random, without replacement, from the set F2 and let S2 denote the set of these chosen elements.

Let G2 = G1 ∪\cup S2. Finally, two elements are chosen at random, without replacement, from the set G2 and let S3 denote the set of these chosen elements.

Let E3 = E2 ∪\cup S3. Given that E1 = E3, let p be the conditional probability of the event S1 = {1, 2}. Then the value of p is

Multiple Choice Options

  • A 15{1 \over 5}
  • B 35{3 \over 5}
  • C 12{1 \over 2}
  • D 25{2 \over 5}

Step-by-Step Verified Solution

JEE Advanced 2021 Paper 1 Online Mathematics - Probability Question 16 English Explanation
To find : Probability P = P(S1∩(E1=E3))P(E1=E3)=P(A1,2)P(A){{P({S_1} \cap ({E_1} = {E_3}))} \over {P({E_1} = {E_3})}} = {{P({A_{1,2}})} \over {P(A)}}

where P(A)=P(A1,2)+P(A1,3)+P(A2,3)P(A) = P({A_{1,2}}) + P({A_{1,3}}) + P({A_{2,3}})

Also, A1, 2 represents 1, 2 chosen at start and similarly others.

Now, P(A1,2)=13×1×3C14C2×15C2=13×36×110=13×12×110P({A_{1,2}}) = {1 \over 3} \times {{1 \times {}^3{C_1}} \over {{}^4{C_2}}} \times {1 \over {{}^5{C_2}}} = {1 \over 3} \times {3 \over 6} \times {1 \over {10}} = {1 \over 3} \times {1 \over 2} \times {1 \over {10}}

P(A1,3)=13×1×2C14C2×15C2=13×23×110P({A_{1,3}}) = {1 \over 3} \times {{1 \times {}^2{C_1}} \over {{}^4{C_2}}} \times {1 \over {{}^5{C_2}}} = {1 \over 3} \times {2 \over 3} \times {1 \over {10}}

P(A2,3)=13×[3C2×14C2×14C2+1×3C14C2×15C2]P({A_{2,3}}) = {1 \over 3} \times \left[ {{{{}^3{C_2} \times 1} \over {{}^4{C_2}}} \times {1 \over {{}^4{C_2}}} + {{1 \times {}^3{C_1}} \over {{}^4{C_2}}} \times {1 \over {{}^5{C_2}}}} \right]

=13×[36×16+36×110]=13[12×16+12×110]= {1 \over 3} \times \left[ {{3 \over 6} \times {1 \over 6} + {3 \over 6} \times {1 \over {10}}} \right] = {1 \over 3}\left[ {{1 \over 2} \times {1 \over 6} + {1 \over 2} \times {1 \over {10}}} \right]

P(A)=13[12×110+12×0+12×110+12×16+23×110+13×0]P(A) = {1 \over 3}\left[ {{1 \over 2} \times {1 \over {10}} + {1 \over 2} \times 0 + {1 \over 2} \times {1 \over {10}} + {1 \over 2} \times {1 \over 6} + {2 \over 3} \times {1 \over {10}} + {1 \over 3} \times 0} \right]

=13[14]=112= {1 \over 3}\left[ {{1 \over 4}} \right] = {1 \over {12}}

∴\therefore Required probability =13(12×110)112= {{{1 \over 3}\left( {{1 \over 2} \times {1 \over {10}}} \right)} \over {{1 \over {12}}}} [ ∵\because P=P(A1,2)P(A)P = {{P({A_{1,2}})} \over {P(A)}} ]

=160×121=15= {1 \over {60}} \times {{12} \over 1} = {1 \over 5}

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