Assignment 3 Solution

Assignment 3 Solution - MATH1040/7040 ASSIGNMENT 3...

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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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