This preview shows pages 1–6. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 1 Probability 1.1 Basic Concepts 1. 1.12 (a) O = & HHHH;HHHT;HHTH;HHTT;HTHH;HTHT;HTTH;HTTT; THHH;THHT;THTH;THTT;TTHH;TTHT;TTTH;TTTT (b) A = f HHHH;HHHT;HHTH;HTHH;THHH g B = f HHTT;HTHT;HTTH;HTTT;THHT;THTH;THTT;TTHH;TTHT;TTTH;TTTT g C = f HHHH;HHHT;HTHH;HTHT;THHH;THHT;TTHH;TTHT g D = f HTTT;THTT;TTHT;TTTH g . Therefore, P ( A ) = 5 16 P ( A \ B ) = P ( ? ) = 0 P ( B ) = 11 16 P ( A \ C ) = P ( f HHHH;HHHT;HTHH;THHH g ) = 4 16 P ( D ) = 4 16 P ( A [ C ) = P ( f HHHH;HHHT;HHTH;HTHH;HTHT;THHH;THHT;TTHH;TTHT g ) = 9 16 P ( B \ D ) = P ( f HTTT;THTT;TTHT;TTTH g ) = 4 16 2. 1.14 & P ( A [ B ) = P ( A ) + P ( B ) P ( A \ B ) = 0 : 4 + 0 : 5 : 3 = 0 : 6 & Because A = ( A \ B ) [ ( A \ B ) , and ( A \ B ) \ ( A \ B ) = ? , we have P ( A ) = P ( A \ B ) + P ( A \ B ) 1 or P ( A \ B ) = P ( A ) & P ( A \ B ) = 0 : 4 & : 3 = 0 : 1 Because A [ B = ( A \ B ) , we have P ( A [ B ) = P & ( A \ B ) = 1 & P ( A \ B ) = 1 & : 3 = 0 : 7 3. 1.15 We have 1 = P ( O ) = P ( A [ B ) = P ( A ) + P ( B ) & P ( A \ B ) Therefore, P ( A \ B ) = P ( A ) + P ( B ) & 1 = 0 : 7 + 0 : 9 & 1 = 0 : 6 4. 1.16 From P ( A [ B ) = P ( A ) + P ( B ) & P ( A \ B ) , we obtain P ( A \ B ) = P ( A ) + P ( B ) & P ( A [ B ) = 0 : 4 + 0 : 5 & : 7 = 0 : 2 From A [ B = ( A \ B ) , we have P ( A [ B ) = P & ( A \ B ) = 1 & P ( A \ B ) = 1 & : 2 = 0 : 8 5. 1.17 b P ( A ) = 2 38 c P ( B ) = 4 38 d P ( D ) = 18 38 2 1.2 Methods of Enumeration 1. 1.21 & 8 8 8 8 = 8 4 = 4096 2. 1.22 & 4 3 2 = 24 3. 1.23 (a) 26 26 10 10 10 10 = 6760000 (b) 26 26 26 10 10 10 = 17576000 4. 1.29 & 9! = 362880 & It is equivalent to &nd the six seatsout of nine seats. Hence, it is & 9 6 = 84 & 2 9 = 512 5. 1.210. Hint: Use ( a + b ) n = P n r =0 & n r a r b n & r & n X r =0 ( 1) r n r = n X r =0 n r ( 1) r (1) n & r = ( 1 + 1) n = 0 n = 0 & n X r =0 n r = n X r =0 n r (1) r (1) n & r = (1 + 1) n = 2 n 3 1.3 Conditional Probability 1. 1.32 (a) P ( A 1 ) = 1041 1456 (b) P ( A 1 j B 1 ) = P ( A 1 \ B 1 ) P ( B 1 ) = 392 = 1456 633 = 1456 = 392 633 (c) P ( A 1 j B 2 ) = P ( A 1 \ B 2 ) P ( B 2 ) = 649 = 1456 823 = 1456 = 649 823 2. 1.34 (a) P ( Two hearts ) = P ( A heart on the second draw \ A heart on the &rst draw ) = P ( A heart on the second draw j A heart on the &rst draw ) P ( A heart on the &rst draw ) If you get a heart in the &rst draw, then you have 12 hearts left in the remaining 51 cards. Therefore, P ( A heart on the second draw j A heart on the &rst draw ) = 12 51 Also, P ( A heart on the &rst draw ) = 13 52 Therefore, the answer is 12 51 13 52 = 1 17 (b) P ( A heart on the &rst draw a club on the second draw ) = P ( A club on the second draw j A heart on the &rst draw ) P ( A heart on the &rst draw ) If the &rst draw was a heart, there are 13 clubs in the remaining 51 cards. Therefore, P ( A club on the second draw j A heart on the &rst draw ) = 13 51 4 AlsoAlso, P ( A heart on the &rst draw...
View Full
Document
 Spring '07
 Guggenberger

Click to edit the document details