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Grade: 12

                        

A is set containing n elements. A subset P of A is chosen at random. The set A reconstructed by replacing the elements of P. A subset Q of A is again chosen at random. Find the probability that P and Q have no common elements. in this what to get n(s)

2 years ago

Answers : (1)

Arun
24741 Points
							
 
 
Please find the answer to your question
Set A has a elements.
∴ Number of subsets of A = 2”
∴ Each one of P and Q can be selected in 2” ways.
Hence total no. of ways of selecting P and Q = 2” = 4”.
Let P contains r elements, where r varies from 0 to n, Then, P can be chosen in n Cr ways.
Now as P ∩ Q = ϕ, Q can be chosen from the set of all subsets of set consisting of remaining (n – r) elements. This can be done in 2n – r ways.
∴ P and Q can be chosen in n C. 2n – r ways. But, r can vary from 0 to n
∴ total number of disjoint sets P and Q are
\sum_{r=0}^{n} ^n C_{r}2^n^-^r= (1 + 2)n = 3n
NOTE THIS STEP:
∴ Required probability = 3n/4n = (3/4)n
ALTERNATE SOLUTION:
Let A = {a1, a2, a3, . . . . . . . . . . ., an}
For each ai, 1 ≤ i n, there Aries 4 cases
(i) a1 ∈ P and a1 ∈ Q
(ii) a1 ∉ P and a1­ ∈ Q
(iii) a1 ∈ P and a1 ∉ Q
(iv) a1 ∉ P and a1 ∉ Q
∴ total no. of ways of choosing P and Q is 4n. Here case (i) is not favorable as P ∩ Q = ϕ
∴ For each element there are 3 favorable cases and hence total no. of favorable cases 3”
Hence prob. (P ∩ Q) = ϕ) = 3n/4n = (3/4)n
ALTERNATE SOLUTION:
The set P be the empty set, or one element set or two elements set . . . . . . . . . or n elements set. Then the set Q will be chosen from amongst the remaining n elements or (n – 1). Element for (n – 2) elements . . . . . . . . . . . . . . . or no elements. Now if P is the empty set then prob. of its choosing is nC0/2n, if it is one element set the then prob. of its choosing is C1/2n, and so on. When the set P consisting of r elements is chosen from A, then the prob. of choosing the set Q from amongst the remaining n - r elements 2n - r /2n. Hence the prob. that P and Q have no common elements is given by
\sum_{r=0}^{n} ^n C_{r/2n . 2n – r/2n = 1/4n \sum_{r=0}^{n} ^n C_{r2n -r
= 1/4n (1 + 2)n (Using Binomial thm.) = 3n/4n = (3/4)n
2 years ago
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