STA3007_0506_t01_ssol

STA3007_0506_t01_ssol - STA 3007 Applied Probability...

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STA 3007 Applied Probability 2005 Tutorial 1 Suggested Solution 1. Review on Probability i. 1.The probability mass function of Z p Z (1) = ( 1 2 )( 1 6 ) = 1 12 p Z (2) = ( 1 2 )( 1 3 ) = 1 6 p Z (3) = ( 1 2 )( 1 2 ) = 1 4 p Z (4) = ( 1 2 )( 1 6 ) = 1 12 p Z (5) = ( 1 2 )( 1 3 ) = 1 6 p Z (6) = ( 1 2 )( 1 2 ) = 1 4 2. Expectation of Z E [ Z ] = 6 z =1 zp Z ( z ) E [ Z ] = 1( 1 12 ) + 2( 1 6 ) + 3( 1 4 ) + 4( 1 12 ) + 5( 1 6 ) + 6( 1 4 ) E [ Z ] = 3 . 833 3. Variance of Z var [ Z ] = E [ Z 2 ] - [ E [ Z ]] 2 E [ Z 2 ] = 6 z =1 z 2 p Z ( z ) E [ Z 2 ] = 1 2 ( 1 12 ) + 2 2 ( 1 6 ) + 3 2 ( 1 4 ) + 4 2 ( 1 12 ) + 5 2 ( 1 6 ) + 6 2 ( 1 4 ) E [ Z 2 ] = 17 . 5 var [ Z ] = 17 . 5 - (3 . 833) 2 var [ Z ] = 2 . 806 ii. Let N be the number of flips required to have the first head appears: Y = ± 1 , if the first flip results in a head 0 , if the first flip results in a tail By using tower expectation: E [ X ] = E Y [ E X [ X | Y ]] E [ X ] = y E X [ X | Y = y ] Pr { Y = y } E [ N ] = y E [ N
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This note was uploaded on 05/21/2011 for the course STA 3007 taught by Professor Kb during the Spring '11 term at CUHK.

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STA3007_0506_t01_ssol - STA 3007 Applied Probability...

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