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Unformatted text preview: MATH1040/7040 ASSIGNMENT 3 SOLUTIONS . 6 .
1. (a) 2—2?=—2x31—2x41—2x51—2x61=
i=3 Hence y=—36 —6—8——lO—12=—36 (b) 22:(—1)kk;(—1)0x0+(41)1x1+(—1)2x2=0—l+2=1
16:0 7 _5 _
(e) Z 7127 = —6m — 533 = —ll:z;
i=~6 ‘
(d) Z 22' = 36, so —4z — 32 — 2z = 36, so —9z = 36
i=—4
Hence z = —4
1 ,
(e) Z ~2x =40, so —2:z; — 2:2: — 2w — 2w —’ 2:1: = 40, so —10z ; 40
i=—3 .
Hence a; = —4
2. (a) 4+8+12+16+20+24 (b) . Prob (1:1 is odd) = Z = =4xl+4x2+4x3+4x4+4x5+4x6
6
= 242'
i=1 2m+l+3x+4+4m+9+5x+16+... ‘
=2x+12+3w+22+4$+32+4m+42+... = 20+ 1):c+'£2 OR Zim+ (2'— 1)2
i=1‘ i=2 2 [oh— . Prob (t1 = 5) = O . Prowl < 2) =2 . l
. Prob (131 IS odd and t1 < 2) = 4—1 . 2 1
. Prob(t11s odd or 751 < 2) = Z = 5 l
. Prob (t1 is odd given that £1 < 2) = I = 1 1 . Prob (t1 is odd) = 5, and Prob (t2 is odd) = : . Now t1 and t2 are chosen independently, so Prob (both t1 and 152 are odd ) = Prob (t1 is odd) >< Prob (t; is odd). Hence Prob (both t1 and 252 are odd ) = 8. By the principle of inclusion\exclusion, Prob (751 is odd or t2 is odd) = Prob (t1 is odd) + Prob (t2 is odd) — Prob (both t1 and t2 are odd )
l l 1 3 Hence Prob (t1 is odd or 152 is odd) = 5 + 5 — Z = Z 9. Now t1 and 752 are chosen independently, so Prob (t1 is odd given that t2 is odd ) = Prob (151 is Odd). Hence Prob (t1 is odd given that 152 is odd ) _= g 4. 1. B =~ {2,7,6, 5, —3}
2. D u A = {3,2,0, —2} u {3,2, 1, 7, 41,4} = {3, 2, 1,0, —2, 7, ~1,4}
3. Am B = {3,2, 1,7, —1,4} m {2, 7,6, 5, —3} = {2,7}~ ‘4. A\D = {3, 2, 1,7, —1,4}\{3, 2, 0, —2} = {1,7, —1,4} 5.
B\ (A u D) = {2, 7, 6, 5, —3}\ ({3, 2, 1, 7, —1, 4} u {3, 2, o, —2})
= {2,7,6,5,—3}\{3,2, 1,0,7, —2, —1,4}
= {6, 5, —3}
6.
(A u B) u D = ({3, 2, 1, 7, —1,4} 0 {2, 7, 6, 5, —3}) U {3, 2, 0, —2}
= {3, 2, 1, 7, —1, 6, 5,4, —3} u {3, 2,0, —2}
={3,2,1,0,7,—2,—1,6,5,4,—3} ‘
7_. B u (D\A) = {2, 7, 6, 5, —3} u ({3, 2,0, —2}\{3, 2, 1,7, —1,4})
= {277,6751_3} U {01—2}
= {2,0, 7, —2,6, 5, —3} s. (Z)\A = 0\{3,2,1,7,—1,4} =0 9. ' (An D>\<B\A> = ({3,2, 1,7, —1,4} 0 {3,2,0 —2}> \ ({2,7,6,5,—3}\{3,2, 1,7,—1,41)
= {3,2}\{6, 5, —3}
= {3,2} ' 10. 5. a. Complete the following table: Contaminated Non Contaminated Total
Positive Test 18,675 ' 2,550 21,225
Negative Test 75 103,700 103,775
106,250 ' 125,000 b. What is the probability that a person randomly selected from the population:
18, 750 (1) P(A) = 125,000 =0.15
21,225 _
(2) 13(3) _ 125,000 _ 0.1698
18,675 _
(3) P(AnB) _ 125,000 _ 0.1494 _ (4) P(A u B) = 0.15 + 0.1698 — 0.1494 = 0.1704 ' 2,550 _'
(5) P(C) _ 125,000 00204
75
6 D = = .
( ) P( ) 125,000 00006 . c. P(E) = 0.0204 + 0.0006 = 0.0210
75 d. P F =
( ) 103,775 = 0.0007...
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