Dr. Hackney STA Solutions pg 8

# Dr. Hackney STA Solutions pg 8 - n odd the proof is...

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Second Edition 1-5 styles. There are two ways of choosing within a given shoe style (left shoe or right shoe), which gives 2 2 r ways of arranging each one of the ( n 2 r ) arrays. The product of this is the numerator ( n 2 r ) 2 2 r . 1.22 a) ( 31 15 )( 29 15 )( 31 15 )( 30 15 ) ··· ( 31 15 ) ( 366 180 ) b) 336 366 335 365 ··· 316 336 ( 366 30 ) . 1.23 P ( same number of heads ) = n X x =0 P (1 st tosses x, 2 nd tosses x ) = n X x =0 " ± n x ²± 1 2 ² x ± 1 2 ² n - x # 2 = ± 1 4 ² n n X x =0 ± n x ² 2 . 1.24 a. P ( A wins) = X i =1 P ( A wins on i th toss) = 1 2 + ± 1 2 ² 2 1 2 + ± 1 2 ² 4 ± 1 2 ² + ··· = X i =0 ± 1 2 ² 2 i +1 = 2 / 3 . b. P ( A wins) = p + (1 - p ) 2 p + (1 - p ) 4 p + ··· = i =0 p (1 - p ) 2 i = p 1 - (1 - p ) 2 . c. d dp ³ p 1 - (1 - p ) 2 ´ = p 2 [1 - (1 - p ) 2 ] 2 > 0. Thus the probability is increasing in p , and the minimum is at zero. Using L’Hˆ opital’s rule we ﬁnd lim p 0 p 1 - (1 - p ) 2 = 1 / 2. 1.25 Enumerating the sample space gives S 0 = { ( B,B ) , ( B,G ) , ( G,B ) , ( G,G ) } ,with each outcome equally likely. Thus P (at least one boy) = 3 / 4 and P (both are boys) = 1 / 4, therefore P ( both are boys | at least one boy ) = 1 / 3 . An ambiguity may arise if order is not acknowledged, the space is S 0 = { ( B,B ) , ( B,G ) , ( G,G ) } , with each outcome equally likely. 1.27 a. For
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Unformatted text preview: n odd the proof is straightforward. There are an even number of terms in the sum (0 , 1 , ,n ), and ( n k ) and ( n n-k ) , which are equal, have opposite signs. Thus, all pairs cancel and the sum is zero. If n is even, use the following identity, which is the basis of Pascals triangle: For k &gt; 0, ( n k ) = ( n-1 k ) + ( n-1 k-1 ) . Then, for n even n X k =0 (-1) k n k = n + n-1 X k =1 (-1) k n k + n n = n + n n + n-1 X k =1 (-1) k n-1 k + n-1 k-1 = n + n n - n-1 - n-1 n-1 = 0 . b. Use the fact that for k &gt; 0, k ( n k ) = n ( n-1 k-1 ) to write n X k =1 k n k = n n X k =1 n-1 k-1 = n n-1 X j =0 n-1 j = n 2 n-1 ....
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## This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.

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