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Consider 'A is a set containing n elements. A subset 'P' of 'A' is chosen at random The set 'A' is reconstructed by replacing the elements of 'A'. A subset 'Q' of 'A' is again chosen at random. What is the probability that 'P' and 'Q' have no common element?
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Suppose you have $3$ containers,$ A, P \ and \ Q$.

Initially, $A$ contains $n$ distinct objects say pens and $P$ and $Q$ are empty.

Now you pick out say k pens from container A and put them in container P.

Now you drop all the pens back to $A$ from $P$.So your container A is reconstructed again.

At this moment your all pens are there in $A$ and $P$ and $Q$ are empty,

Now you again pick $r$ pens from $A$ and put them in container $Q.$

$\text{So, there are 4 possibilities for each pen-}$
1. Getting selected for both P and Q.
2. Getting selected for P only.
3.Getting selected for Q only.
4. Getting selected for none.

Out of these, 2,3 and 4 are our favourable cases.
So  required probability = $\frac{3^n}{4^n}$
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Suppose we have n different elements say (x1,x2,x3,....................xn) and two subsets P and Q.

now either the chosen element will belong to p or not or it may belong to q or not.

means               1) x1 belongs to both P and Q.

                           2)  to only P not Q.

                           3 to only Q not P.

                           4) none of P and Q.

so an element can be placed  in 4 different ways in two sets.

so for n elements it can be done in 4n ways(total ways).

favourable outcomes = P ∩ Q = ϕ , which can be selected in 3 ways (i.e. option 2 ,3 or 4).

for n elements it can be done in 3n ways

probability = $\large \frac{3^n}{4^n}$  = $\large (\frac{3}{4})^n$

                

 

 

 

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