08Ma2bHw1Sol

08Ma2bHw1Sol - MA2B HW 1 SOLUTIONS Problem 1 1.3.8 Solution...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
MA2B HW 1 SOLUTIONS Problem 1: 1.3.8 Solution: For this problem, we use the rules of probability as stated earlier in this chapter. (a) P ( A B ) = P ( A ) + P ( B ) - P ( AB ) = 0 . 6 + 0 . 4 - 0 . 2 = 0 . 8 . (b) P ( A c ) = 1 - P ( A ) = 1 - 0 . 6 = 0 . 4 . (c) P ( B c ) = 1 - P ( B ) = 1 - 0 . 4 = 0 . 6 . (d) Note that we can partition B into AB and A c B , so by the rule of addition, P ( B ) = P ( AB ) + P ( A c B ) which implies P ( A c B ) = P ( B ) - P ( AB ) = 0 . 4 - 0 . 2 = 0 . 2 . (e) First, by the same method as above, as A can be partitioned into AB and AB c , we find that P ( AB c ) = P ( A ) - P ( AB ) = 0 . 6 - 0 . 2 = 0 . 4 . Thus, using inclusion-exclusion: P ( A B c ) = P ( A ) + P ( B c ) - P ( AB c ) = 0 . 6 + 0 . 6 - 0 . 4 = 0 . 8 . (f) Again, by partitioning A c into A c B and A c B c , we obtain: P ( A c B c ) = P ( A c ) - P ( A c B ) = 0 . 4 - 0 . 2 = 0 . 2 . Problem 2: 1.3.13 Solution: First, note that this is trivially true for n = 1 (As clearly P ( A 1 ) P ( A 1 )), so we first prove the case n = 2. Then, by inclusion-exclusion, we have: P ( 2 [ i =1 A i ) = P ( A 1 A 2 ) = P ( A 1 ) + P ( A 2 ) - P ( A 1 A 2 ) . However, by non-negativity, we know P ( A 1 A 2 ) 0, implying that P ( 2 [ i =1 A i ) P ( A 1 ) + P ( A 2 ) . So, now we assume that the statement is true for n = k 2. Then, we have: P ( k +1 [ i =1 A i ) = P ( A k +1 ( k [ i =1 A i )) P ( A k +1 ) + P ( k [ i =1 A i ) ( * ) 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
2 MA2B HW 1 SOLUTIONS P ( k +1 [ i =1 A i ) P ( A k +1 ) + k X i =1 P ( A i ) = k +1 X i =1 P ( A i ) ( ** ) ( statement ( * ) is true by the truth of the statement for n = 2, and statement ( ** ) is true by the induction hypothesis). Thus, P ( S k +1 i =1 A i ) k +1 i =1 P ( A i ), so by induction, for any natural number n , it is true that P ( n [ i =1 A i ) n X i =1 P ( A i ) . Problem 3: 1.4.5 Solution: (a) Tree diagram omitted at this time. (b) Along the branches of the tree diagram, if we let
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/08/2011 for the course MA 2b taught by Professor Makarov,n during the Winter '08 term at Caltech.

Page1 / 5

08Ma2bHw1Sol - MA2B HW 1 SOLUTIONS Problem 1 1.3.8 Solution...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online