JEE MAIN - Mathematics (2021 - 24th February Evening Shift - No. 9)
The probability that two randomly selected subsets of the set {1, 2, 3, 4, 5} have exactly two elements in their intersection, is :
$${{135} \over {{2^9}}}$$
$${{65} \over {{2^8}}}$$
$${{65} \over {{2^7}}}$$
$${{35} \over {{2^7}}}$$
Explanation
Given, set P = {1, 2, 3, 4, 5}
Let the two subsets be A and B
Then, n (A $$\cap$$ B) = 2 (as given in question)
We can choose two elements from set P in 5C2 ways.
After choosing two common elements for set A and B, each of remaining three elements from set P have three choice (1) It can go to set A (2) It can go to set B (3) It don't go to any sets it stays at set P.
$$ \therefore $$ Total ways for the three elements = 3 $$ \times $$ 3 $$ \times $$ 3 = 33
$$\therefore$$ Required probability = $${{{}^5{C_2} \times {3^3}} \over {\left( {{2^5}} \right)\left( {{2^5}} \right)}}$$ = $${{{}^5{C_2} \times {3^3}} \over {{4^5}}} = {{10 \times 27} \over {{2^{10}}}} = {{135} \over {{2^9}}}$$
Let the two subsets be A and B
Then, n (A $$\cap$$ B) = 2 (as given in question)
We can choose two elements from set P in 5C2 ways.
After choosing two common elements for set A and B, each of remaining three elements from set P have three choice (1) It can go to set A (2) It can go to set B (3) It don't go to any sets it stays at set P.
$$ \therefore $$ Total ways for the three elements = 3 $$ \times $$ 3 $$ \times $$ 3 = 33
$$\therefore$$ Required probability = $${{{}^5{C_2} \times {3^3}} \over {\left( {{2^5}} \right)\left( {{2^5}} \right)}}$$ = $${{{}^5{C_2} \times {3^3}} \over {{4^5}}} = {{10 \times 27} \over {{2^{10}}}} = {{135} \over {{2^9}}}$$
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