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solution_set1 - EE 511 Solutions to Problem Set 1 1(i A A =...

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EE 511 Solutions to Problem Set 1 1. (i) A + A = S and A A = φ . Therefore, P ( A ) + P ( A ) = P ( S ) = 1 and P ( A ) = 1 - P ( A ). (ii) P ( A ) 0. Therefore, P ( A ) 1. (iii) φ + S = S and φS = φ . Therefore, P ( φ ) + P ( S ) = P ( S ) and P ( φ ) = 0. (iv) B = BS = B ( A 1 + ... + A n ). Since BA i and BA j are disjoint for i 6 = j , P ( B ) = P ( BA 1 ) + P ( BA 2 ) + ... + P ( BA n ). 2. B = A + AB where A ( AB ) = φ . Therefore, P ( A ) + P ( AB ) = P ( B ). Since P ( AB ) 0, P ( A ) P ( B ). 3. A + B = A + AB with A ( AB ) = φ . Therefore, P ( A + B ) = P ( A ) + P ( AB ). Similarly, B = ( A + A ) B = AB + AB . Therefore, P ( B ) = P ( AB ) + P ( AB ). Substituting this in the equation for P ( A + B ), we get P ( A + B ) = P ( A ) + P ( B ) - P ( AB ) Now, P(A+B+C) = P((A+B)+C) = P(A+B) + P(C) - P((A+B)C) = P(A) + P(B) - P(AB) + P(C) - P(AC + BC) = P(A) + P(B) + P(C) - P(AB) - (P(AC) + P(BC) - P(ACBC)) = P(A) + P(B) + P(C) - P(AB) - P(AC) - P(BC) + P(ABC) 4. We want to show P ( N X i =1 A i ) N X i =1 P ( A i ). This can be done in several ways. Solution 1: We have shown in problem 3 that P ( A + B ) = P ( A ) + P ( B ) - P ( AB ), i.e., P ( A + B ) P ( A ) + P ( B ). Using this result repeatedly, we get P ( N X i =1 A i ) = P ( A 1 + N X i =2 A i ) P ( A 1 ) + P ( N X i =2 A i ) P ( N X i =2 A i ) = P ( A 2 + N X i =3 A i ) P ( A 2 ) + P ( N X i =3 A i ) .
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