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# DM - Grand Test -Q7

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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 , 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)

reshown Aug 31, 2020

+1 vote

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 (11,240 points)
selected Sep 29, 2020