MDM4UI Unit 1 Complete Solutions-Akter

MDM4UI Unit 1 Complete Solutions-Akter - Solutions to Unit...

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Solutions to Unit 1.3 1a) { } 7 of diamonds 1b) { } ace of spades, ace of diamonds, ace of hearts, ace of clubs 1c) { } all of clubs 2, 3, 4, 5, 6, 7, 8, 9,10 1d) { of clubs, diamonds, hearts, and spades } 2, 4, 6, 8,10 3a) There are 5 possible outcomes 3b) ( ) ( ) ( ) black black all possible 3 5 n P n = = 3c) ( ) ( ) ( ) red red all possible 2 5 n P n = = 3d) There are no possible outcomes therefore the probability is 0 4a) ( ) ( ) ( ) joker joker all possible cards 2 54 1 27 n P n = = = 4b) ( ) ( ) ( ) red joker red joker all possible cards 1 54 n P n = =
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4c) ( ) ( ) ( ) queen queen all possible cards 4 54 2 27 n P n = = = 4d) ( ) ( ) ( ) black black all possible cards 27 54 1 2 n P n = = = 4e) ( ) ( ) ( ) <10 <10 all possible cards 36 54 2 3 n P n = = = 4f) ( ) ( ) ( ) red red all possible cards 3 54 1 18 n P n = = = 5a) ( ) ( ) ( ) tail tail all possible cards 1 2 n P n = = 5b) ( ) ( ) ( ) 3 3 all possible cards 1 6 n P n = =
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5c) ( ) ( ) ( ) red red all possible cards 26 52 1 2 n P n = = = 5d) ( ) ( ) ( ) B7 B7 all possible cards 2 54 1 27 n P n = = = 5e) ( ) ( ) ( ) <jack <jack all possible cards 40 52 10 13 n P n = = =
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Unit 1.4 Solutions 6. a) ( ) ( ) ( ) ( ) ( ) odd number any number 4 8 n A n S n n = = = P A b) ( ) ( ) ( ) ( ) ( ) divisible by 4 any number 2 8 1 4 n A n S n n = = = = P A c) ( ) ( ) ( ) ( ) ( ) <3 any number 2 8 1 4 n A n S n n = = = = P A 7. a) ( ) ( ) ( ) ( ) ( ) red blocks number blocks 3 12 1 4 n A P A n S n n = = = =
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b) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 red blocks 1 number blocks 3 1 12 3 4 P A P A n A n S n n = = = = = 8. a) ( ) ( ) ( ) ( ) ( ) S's letters 1 11 n A P A n S n n = = = b) ( ) ( ) ( ) ( ) ( ) M's letters 2 11 n A n S n n = = = P A c) ( ) ( ) ( ) ( ) ( ) vowels letters 4 11 n A n S n n = = = P A
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10 Possible values {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} 10. a) ( ) ( ) ( ) ( ) ( ) all heads possible values 1 8 n A P A n S n n = = = 10 b) ( ) ( ) ( ) ( ) ( ) 1 tail, 2 tail, 3 tail possible values 7 8 n A P A n S n n = = = 10 c) ( ) ( ) ( ) ( ) ( ) HHT, HTH, THH possible values 3 8 n A P A n S n n = = = 12. Die 1 2 3 4 5 6 1 1, 1 2, 1 3, 1 4, 1 5, 1 6, 1 2 1, 2 2, 2 3, 2 4, 2 5, 2 6, 2 3 1, 3 2, 3 3, 3 4, 3 5, 3 6, 3 4 1, 4 2, 4 3, 4 4, 4 5, 4 6, 4 5 1, 5 2, 5 3, 5 4, 5 5, 5 6, 5 6 1, 6 2, 6 3, 6 4, 6 5, 6 6, 6 12 a) ( ) 36 n S =
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12 b) ( ) ( ) ( ) ( ) ( ) totals 7 possible values 6 36 1 6 n A P A n S n n = = = = 12 c) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 totals 7 1 possible values 6 1 36 5 6 P A P A n A n S n n ′ = = = = = 14. 44 100 = 44% this will be equal to the probability ratio we need let the number of caffeine-free diets be x , ( ) ( ) diet 44 100 all drinks 44 16 100 44 1936 44 1600 100 336 56 6 n n x x x x x x = + = + + = + = =
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Unit 4.1 Solutions 1. a) {6, 9} b) {5, 2, 6, 9, 12, 10} c) {9,10} d) {4, 2, 9, 10, 6, 1} e) f) {9} 6. a) 3 7 10 T C F S b) ( ) ( ) ( ) 10 20 1 2 n C n S = = = P C c) ( ) ( ) ( ) 3 20 n T n S = = P C
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d) ( ) ( ( ) ( ) ) 1 1 7 1 20 13 20 P F n F n S ′ = = = = P F 8 16 14 12 S C HA 8. The number combined in both sets is 50-12=38 Let HA represent Head aches only, B represent both, and C represent colds Therefore HA+B=30 C+B=24 HA+B+C=38 We have three equations and three variables, therefore we can do this. Rewrite the first two equations in terms of B HA=30-B C=24-B Plug these into the third equation. (30-B)+B+(24-B)=38 54-B=38 B=16 Therefore HA 30 16 14 = = 24 16 8 C = =
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b i) 14 students have just a headache ( ) ( ) ( ) 14 50 7 25 n HA P HA n S = = = b ii) 14+8+16=38 have headache or cold ( ) ( ) ( ) or C or C 38 50 19 25 n HA P HA n S = = = b iii) 12+14=26 have no cold symptom ( ) ( ) ( ) none none 26 50 13 25 n P n S = = =
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Unit 1.6 Solutions 2a) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0.7 0.5 0.2 0.7 0.7 0 P A B P A P B P A B P A B P A B P A B = + = + = = Since P A =0, mutually exclusive.
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