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Normalization Lectures


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DM - Grand Test -Q7


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+2 votes

Let S be a set with six elements. Let P be the set of all subsets of S. Subsets A and B of S, not necessarily distinct, are chosen independently and at random

from P. The probability that B is contained in at least one of A or S\A is m/n^r , where m, n,and r are positive integers, n is prime, and m and n are relatively prime. What is the value of m + n + r ?

(The set S\A is the set of all elements of S which are not in A)

asked Aug 30, 2020 by gbeditor-2 (72,840 points)
reshown Aug 31, 2020 by gbeditor-2

2 Answers

+1 vote
 
Best answer

n(A) = 2^{6}

n(B) = 2^{6}

Sample space : n(A)*n(B) =  2^{6}*2^{6} = 2^{12}

 

CASE 1 :  B is contained in A

^{6}C_{0}*2^{6} + ^{6}C_{1}*2^{5} + ^{6}C_{2}*2^{4} +... + ^{6}C_{6}*2^{0} = (1+2)^{6} = 3^{6}

 

Case 2 : B contained in A^{'} (or S-A)

But as B\subset A so this case is same as Case 1 i.e. 3^{6} 

 

Case 3: B is present in Both (we need to subtract this) But as we know if B is present in A and at the same time it is present in A' that means it is null set. so it is sam as B is empty set. 

^{6}C_{0}*2^{6}

CASE 1 + CASE 2 - CASE 3 is what we wanted.

\frac{2*3^{6} - 2^{6}}{2^{12}}

which reduces to

\frac{697}{2^{11}}

697 + 2 + 11 = 710

 

answered Sep 29, 2020 by tsshashankrustagi (11,240 points)
selected Sep 29, 2020 by deepak-gatebook
0 votes
See solutions below :

https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_10
answered Sep 9, 2020 by deepak-gatebook (226,240 points)
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