By Aitken R.J.

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**Extra info for A Statistical Study of the Visual Double Stars in the Northern Sky (1915)(en)(5s)**

**Example text**

108 21. (a) E[X] since whereas the bus driver selected is equally likely to be from any of the 4 buses, the student selected is more likely to have come from a bus carrying a large number of students. 28 E[Y] = (40 + 33 + 25 + 50)/4 = 37 22. Let N denote the number of games played. (a) E(N) = 2[p2 + (1 − p)2] + 3[2p(1 − p)] = 2 + 2p(1 − p) The final equality could also have been obtained by using that N = 2 + ] where I is 0 if two games are played and 1 if three are played. Differentiation yields that d E[ N ] = 2 − 4 p dp and so the minimum occurs when 2 − 4p = 0 or p = 1/2.

2n (e) Since the 2 players in game i are equally likely to be any of the pairs it follows that 2 2n P(Bi) = 1 . 2 (f) Since the events Bi are mutually exclusive P(∪ Bi) = ∑ P( B ) = (2 i n 2n − 1) = (1 / 2) n −1 2 81. 1 − (9 / 11)15 1 − (9 / 11)30 82. (a) P(A) = P12 + 1 − P12 1 − P22 P( A) or P(A ) = ( )([ ) ] P12 P12 + P22 − P12 P22 (c) similar to (a) with Pi 3 replacing Pi 2 . Chapter 3 37 (b) and (d) Let Pij ( Pij ) denote the probability that A wins when A needs i more and B needs j more and A(B) is to flip.

Not true. Let F = E1. 29. P{next m headsfirst n heads} = P{first n + m are heads}/P(first n heads} 1 1 n +1 n+m = p dp . p n dp = n + m +1 0 0 ∫ Chapter 3 ∫ 45 Chapter 4 Problems 1. 2. 4 2 6 P{X = 4} = = 14 91 2 2 2 1 P{X = 0} = = 14 91 2 4 2 2 1 8 P{X = 2} = = 14 91 2 8 2 1 1 16 P{X = −1} = = 91 14 2 4 8 1 1 32 P{X = 1} = = 91 14 2 8 2 28 P{X = −2} = = 14 91 2 p(1) = 1/36 p(5) = 2/36 p(9) = 1/36 p(15) = 2/36 p(24) = 2/36 p(2) = 2/36 p(6) = 4/36 p(10) = 2/36 p(16) = 1/36 p(25) = 1/36 p(3) = 2/36 p(7) = 0 p(11) = 0 p(18) = 2/36 p(30) = 2/36 p(4) = 3/36 p(8) = 2/36 p(12) = 4/36 p(20) = 2/36 p(36) = 1/36 5 5 5 5 45 5 = = , P{X = 3} = , 10 9 18 10 9 8 36 5 4 3 5 10 5⋅ 4⋅3⋅ 2 5 5 = , P{X = 5} = = P{X = 4} = , 10 9 8 7 168 10 ⋅ 9 ⋅ 8 ⋅ 7 6 252 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1 1 = P{X = 6} = 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 252 4.