f2011sollectexamp14

# f2011sollectexamp14 - 6 In the table below in each cell I...

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Solutions to Chapter 1 Lecture Examples FALL 2011 1. (a) P ( A )=0 . 25 (b) P ( B . 50 (c) P ( C . 75 2. (a) Call the probabilities 9 u , 3 u , 3 u and u .T h e s em u s t s u m t oo n e h u s , 16 u =1 ;hence u / 16 .Thu s ,the probabilities of a, b, c and d are 9 / 16 , 3 / 16 , 3 / 16 and 1 / 16 . (b) P ( A )=9 / 16 ; P ( B )=1 2 / 16 = 0 . 75 ;and P ( C )=7 / 16 . 3. (a) P ( A or B . 40 + 0 . 25 = 0 . 65 . (b) P ( A c - 0 . 40 = 0 . 60 . (c) P ( B c - 0 . 25 = 0 . 75 . 4. P ( A or B . 25 + 0 . 45 - 0 . 20 = 0 . 50 . 5. (a) P ( A . 20 ; P ( B . 55 ;a n d P ( AB . 25 :Th isv io la tesRu le5 ; AB is a subset of A ,soitsprobability cannot be larger. (b) P ( A . 60 ; P ( B . 55 A and B are disjoint: This violates Rule 2. By Rule 3, P ( A or B . 60 + 0 . 55 = 1 . 15 ,whichistoolarge . 6. In the table below, in each cell I have placed the value of X = X 1 + X 2 .Fo rex amp l e , in the center cell, corresponding to X 1 =2 and X 2 ,Ihaveplaced 2+2=4 . X 2 X 1 1 2 3 1 2 3 4 2 3 4 5 3 4 5 6 The nine cells are equally likely, so prob- abilities are obtained by simply counting, yielding the following table. x : 23456 P ( X = x ) : 1 / 92 / 93 / / 91 / 9 7. In the table below, in each cell I have placed the value of X = X 1 X 2 .F o re x am p l e ,i n the cell corresponding to X 1 and X 2 = 3 2 × 3=6 . X 2 X 1 0 1 2 3 4 0 0 0 0 0 0 1 0 1 2 3 4 2 0 2 4 6 8 3 0 3 6 9 12 4 0 4 8 12 16 The 25 cells are equally likely, so probabili- ties are obtained by simply counting, yield- ing the following table. xP ( X = x ) ( X = x ) 00 . 3 6 60 . 0 8 10 . 0 4 80 . 0 8 20 . 0 8 90 . 0 4 30 . 0 8 12 0.08 40 . 1 2 16 0.04 1

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Solutions to Chap. 1 Lecture Examples: Cont. 8. Because the outcomes of each trial are not equally likely, the cells in the table are not equally likely. Instead, we obtain the prob- ability of each cell by using the multiplica- tion rule. For example, P ( X 1 =2 ,X 2 = 3) = 0 . 1(0 . 7) = 0 . 07 .The sep robab i l i t ie s are presented below. X 2 X 1 1 2 3 1 0.04 0.02 0.14 2 0.02 0.01 0.07 3 0.14 0.07 0.49 The values of X = X 1 + X 2 for these nine cells were given above in the solution to ex- ample 6. Using these two tables and adding probabilities, one gets the following distri- bution for X . x : 234 P ( X = x ) : 0.04 0.04 0.29 x : 56 P ( X = x ) : 0.14 0.49 9. The possible values of X are: 3, 4, 5, . .., 9. P ( X =3)= P (1 , 1 , 1) = (0 . 2) 3 =0 . 008 . P ( X =4)=3 P (1 , 1 , 2) = 3(0 . 2) 2 (0 . 1) = 0 . 012 . P ( X =5)=3 P (1 , 1 , 3) + 3 P (1 , 2 , 2) = 3(0 . 2) 2 (0 . 7) + 3(0 . 2)(0 . 1) 2 . 090 . P ( X =6)=6 P (1 , 2 , 3) + P (2 , 2 , 2) = 6(0 . 2)(0 . 1)(0 . 7) + (0 . 1) 3 . 085 . P ( X =7)=3 P (1 , 3 , 3) + 3 P (2 , 2 , 3) = 3(0 . 2)(0 . 7) 2 +3(0 . 1) 2 (0 . 7) = 0 . 315 . P ( X =8)=3 P (2 , 3 , 3) = 3(0 . 1)(0 . 7) 2 = 0 . 147 . P ( X =9)= P (3 , 3 , 3) = (0 . 7) 3 . 343 . 2
Solutions to Chapter 2 Lecture Examples FALL 2011 1. (a) ppqp =(0 . 65) 3 (0 . 35) = 0 . 0961 . (b) (5! / 4!1!)(0 . 65) 4 (0 . 35) = 0 . 3124 . (c) Each day is a trial, with Brad occur- ring labeled an S .F rom(b ) , P ( S )= 0 . 3124 . i. P ( SSFF ppqq = (0 . 3124) 2 (0 . 6876) 2 =0 . 0461 . ii. (4! / 2!2!)(0 . 3124) 2 (0 . 6876) 2 = 0 . 2786 . (d) Let X 1 denote the number of free throws Anna makes on Friday and X 2 denote the number she makes on Sat- urday. Y = X 1 + X 2 . i. P ( Y =2)= P ( X 1 =1 ,X 2 =1)+ P ( X 1 =2 2 =0) . Now, the ±rst of these is 4! 1!3! (0 . 65)(0 . 35) 3 [0 . 65] = 0 . 1115[0 . 65] = 0 . 0725 . The second of these is 4! 2!2! (0 . 65) 2 (0 . 35) 2 [(0 . 35) 2 ]= 0 . 3105(0 . 1225) = 0 . 0380 . Thus, P ( Y )=0 . 0725 + 0 . 0380 = 0 . 1105 . ii. P ( Y =6)= P (4 , 2) + P (3 , 3) . The ±rst of these is (0 . 65) 4 4! 2!2! (0 . 65) 2 (0 . 35) 2 = 0 . 1785(0 . 3105) = 0 . 0554 . The second of these is 4!

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f2011sollectexamp14 - 6 In the table below in each cell I...

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