JEE Advance - Mathematics (2021 - Paper 1 Online - No. 3)
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
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
$${1 \over 5}$$
$${3 \over 5}$$
$${1 \over 2}$$
$${2 \over 5}$$
Explanation
To find : Probability P = $${{P({S_1} \cap ({E_1} = {E_3}))} \over {P({E_1} = {E_3})}} = {{P({A_{1,2}})} \over {P(A)}}$$
where $$P(A) = P({A_{1,2}}) + P({A_{1,3}}) + P({A_{2,3}})$$
Also, A1, 2 represents 1, 2 chosen at start and similarly others.
Now, $$P({A_{1,2}}) = {1 \over 3} \times {{1 \times {}^3{C_1}} \over {{}^4{C_2}}} \times {1 \over {{}^5{C_2}}} = {1 \over 3} \times {3 \over 6} \times {1 \over {10}} = {1 \over 3} \times {1 \over 2} \times {1 \over {10}}$$
$$P({A_{1,3}}) = {1 \over 3} \times {{1 \times {}^2{C_1}} \over {{}^4{C_2}}} \times {1 \over {{}^5{C_2}}} = {1 \over 3} \times {2 \over 3} \times {1 \over {10}}$$
$$P({A_{2,3}}) = {1 \over 3} \times \left[ {{{{}^3{C_2} \times 1} \over {{}^4{C_2}}} \times {1 \over {{}^4{C_2}}} + {{1 \times {}^3{C_1}} \over {{}^4{C_2}}} \times {1 \over {{}^5{C_2}}}} \right]$$
$$ = {1 \over 3} \times \left[ {{3 \over 6} \times {1 \over 6} + {3 \over 6} \times {1 \over {10}}} \right] = {1 \over 3}\left[ {{1 \over 2} \times {1 \over 6} + {1 \over 2} \times {1 \over {10}}} \right]$$
$$P(A) = {1 \over 3}\left[ {{1 \over 2} \times {1 \over {10}} + {1 \over 2} \times 0 + {1 \over 2} \times {1 \over {10}} + {1 \over 2} \times {1 \over 6} + {2 \over 3} \times {1 \over {10}} + {1 \over 3} \times 0} \right]$$
$$ = {1 \over 3}\left[ {{1 \over 4}} \right] = {1 \over {12}}$$
$$\therefore$$ Required probability $$ = {{{1 \over 3}\left( {{1 \over 2} \times {1 \over {10}}} \right)} \over {{1 \over {12}}}}$$ [$$\because$$ $$P = {{P({A_{1,2}})} \over {P(A)}}$$]
$$ = {1 \over {60}} \times {{12} \over 1} = {1 \over 5}$$
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