Consider a set P containing n elements. A subset A of P is drawn and there after set P is reconstructed. Now one
more subset B of P is drawn. Probability of drawing sets A and B so that $A \cap B$ has exactly one element is

Question

Infinity Learn · Accepted Answer

Let xi be any element of set P, we have following $$possibilities1$$ xi∈A,xi∈$$B2$$ xi∈A,xi∉$$B3$$ xi∉A,xi∈$$B4$$ xi∉A,xi∉BClearly, the element    A o B it it belongs to A and B both.Thus out of these $$4$$ ways only first way is favorable. Now the element that we want to be in the intersection can be chosenin'n' different ways. Hence required probability is n⋅$$(3/4)n$$−$$1$$

Consider a set P containing n elements. A subset A of P is drawn and there after set P is reconstructed. Now one
more subset B of P is drawn. Probability of drawing sets A and B so that $A \cap B$ has exactly one element is

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Consider a set P containing n elements. A subset A of P is drawn and there after set P is reconstructed. Now onemore subset B of P is drawn. Probability of drawing sets A and B so that A∩B has exactly one element is

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Ready to Test Your Skills?

free study material

Consider a set P containing n elements. A subset A of P is drawn and there after set P is reconstructed. Now one
more subset B of P is drawn. Probability of drawing sets A and B so that $A \cap B$ has exactly one element is