Econ41-Homework-Hint

# Econ41-Homework-Hint - 1 Probability 1.1 Basic Concepts 1...

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Unformatted text preview: 1 Probability 1.1 Basic Concepts 1. 1.1-2 (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. 1.1-4 & 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. 1.1-5 ¡ 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. 1.1-6 ¡ 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. 1.1-7 b P ( A ) = 2 38 c P ( B ) = 4 38 d P ( D ) = 18 38 2 1.2 Methods of Enumeration 1. 1.2-1 & 8 ¡ 8 ¡ 8 ¡ 8 = 8 4 = 4096 2. 1.2-2 & 4 ¡ 3 ¡ 2 = 24 3. 1.2-3 (a) 26 ¡ 26 ¡ 10 ¡ 10 ¡ 10 ¡ 10 = 6760000 (b) 26 ¡ 26 ¡ 26 ¡ 10 ¡ 10 ¡ 10 = 17576000 4. 1.2-9 & 9! = 362880 & It is equivalent to &nd the six ¡seats¢out of nine seats. Hence, it is & 9 6 ¡ = 84 & 2 9 = 512 5. 1.2-10. 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. 1.3-2 (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. 1.3-4 (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...
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Econ41-Homework-Hint - 1 Probability 1.1 Basic Concepts 1...

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