2. Probability Solutions Updated (RVHS)

River Valley High School, Mathematics Department 2023 Probability T1 Probability 1(a)(i) P(all three cards even) 10 9 8 220 19 18 19  or 10 3 20 3 219             (ii) P(exactly one even) 10 10 9 1520 19 18 38 3 1        or 10 10 1 2 20 3 1538                 (b)(i) 16 16 154 4 520 19 20 19P B    (ii) 15 16 791 420 19 20 19 380 P( ) P(1st card 5 & 2nd card 5) P(1st card 5 & 2nd card 5) P(1st card 5 & 2nd card 5) A B             (iii)   79P( ) 380 79951P 4 P A B AB A    2(a) 16 3)( 4 11 )( 4 1 ''    ABPABP 048 7 16 3 3 1)()()( '  ABPBPABP Hence A and B are not mutually exclusive. (b)(i) P(score is 5) = P[(3,2), (1,1,3), (1,2,2), (1,3,1), (2,1,2), (2,2,1)] = 216 11 216 5 36 1  (ii)  216 1),1,1,1()3( PscoreP 6 3 1,2,3,4,5,6 1,2,4,5 6 1,2,3,4,5 1,2,3,4,5,6

River Valley High School, Mathematics Department 2023 Probability T2 P(score 4) = P[(3,1), (1,1,2),(1,2,1), (2,1,1) ] = 216 9       5 5454 score scorescorePscorescoreP       5 4 score scoreP = 7 3 1191 9  3(i) 0.9 100 – 0.1 20p p   (p)(0.9) + (100 – p)(0.1) = 20 0.9p + 10 – 0.1p = 20 p = 12.5 The proportion of the residents infected is 12.5 1 100 8 or 12.5 % (ii) P( has disease | tested negative ) = P(has disease & tested negative) P(tested negative) 0.1 0.1 0.9(100 ) q q q   900 8 q q  (iii) As the proportion of people getting infected (q) increases, the probability that a person has the disease given that he has been tested negative also increases as seen in the graph. So, the test is not effective. 4(i) P (it will take exactly n throws of the biased die to obtain a ‘6’) = P (the first (n – 1) throws are not ‘6’ and the nth throw is a ‘6’) = 1(1 ) np p (ii) P (it takes exactly 3 throws to obtain a ‘6’) = 2(1 )p p = 9 64 3 2 92 0 64 0.25,0.4243,1.326 (from GC-Polynomial solver) p p p p     

River Valley High School, Mathematics Department 2023 Probability T3 0.4243 0.424p  since 1 14 p  P ( obtain a ‘5’) = 1 0.42430.115145   P (They obtained the same number | they obtained a number larger than 4) P ( they obtained the same number and each obtained a number greater than 4) P ( they each obtained a number greater than 4) =   2 P (each obtained 5 or each obtained 6) P(5, 5) + P(6, 6) P(each obtained 5 or 6) P (5 or 6)  = 2 2 2 (0.11514) (0.4243) (0.11514 0.4243)   = 0.664 (to 3 sf) Alternative solution : = P (each obtained 5 or each obtained 6) P(5, 5) + P(6, 6) P(each obtained a number greater than 4)P (5, 5) + P (6, 6) + 2 P (6, 5) = 2 2 2 2 (0.11514) (0.4243) (0.11514) (0.4243) 2(0.11514)(0.4243)    5(i) P(A wins a race) = 0.2(0.9)+0.8(0.5)=0.58 (ii) P(A wins no more than twice) = 1-P(A wins all 3 matches) = 1- 30.58=0.805 (iii) P(A wins competition|A wins 1st race) = P(A wins competition A wins 1st race) P(A wins 1st race)  P(A wins competition A wins 1st race)  = P(WWW) + P(WWLW) + P(WLWW) + P(WLLWW) + P(WLWLW) + P(WWLLW) =       3 3 3 20.58 0.58 (0.42)2 0.58 (0.42) 3  = 0.462 Required probability = 0.4620.58 = 0.797 6 (i) Let U and R be the events that United and Rover scored a goal with penalty kick respectively. P(match is still undecided after 1 round )

River Valley High School, Mathematics Department 2023 Probability T4 = P(R U ) + P(R’U’) = 0.8  0.9 + 0.2  0.1 = 0.74 = 50 37 P(United won in less than 3 rounds| United scores a goal in the first round) = roundfirsttheingoalascoresUnitedP roundfirsttheingoalascoresUnitedandroundsthanlessinwonUnitedP ( )3( = (0.8 0.1) (0.8 0.9 0.8 0.1) 0.8      = 0.172 (ii) Let X be the random variable for the number of rounds played P(match is decided in at most n rounds) > 0.98 P(X )n > 0.98 P(X = 1) + P(X = 2) + P(X = 3) + ……+P(X = n) > 0.98 Hence 50 13+50 37(50 13)+ 2)50 37( 50 13+……+( 1)50 37 n 50 13> 0.98…… (**) 50 13          12 50 37........50 37 50 371 n > 0.98 n    50 37 < 0.02 Hence n > 50 37ln 02.0ln = 12.992 Least n = 13 Alternative: P(match is decided in at most n rounds) > 0.98 1 – P(match is decided in more than n rounds) > 0.98 1 – P(match is undecided in the first n rounds) > 0.98 1 - n    50 37 > 0.98 7(a)(i) P(A) = 9 1 36 4 P(B) = 4 1 36 9 P(AB) = 1 36 Since P(AB) = P(A) P(B), they are independent. (ii) AB represents the event the card taken is either blue or numbered 1. P(AB) = P(A) + P(B) − P(AB) = 9 4 1 36 36 36  = 1 3

River Valley High School, Mathematics Department 2023 Probability T5 (iii) ' 'P( ' | ') ( ') P A BA B P B  = 11 332 36  = 3 4 (b) Method 1: Required probability =3 2 31 35 34 595   or 0.00504 Method 2: Required probability =4 3 2 3 936 35 34 595    Method 3: Required probability = 4 1 1 1 3 3! 93 36 35 34 595                Method 4: Required probability = 49 3 3 36 595 3        (c) Method 1: Required probability = 32 20 36 20 52 1683 C C  = 0.030897 0.0309 Method 2: Required probability =32 31 30 29 28 17 16 15 14 13...36 35 34 33 32 21 20 19 18 17          = 16 15 14 13 36 35 34 33   = 0.030897 0.0309 8(a) P(obtain a ‘6’) = P(R6) + P(B6) = 1 1 1x x 23 6 9 (b) P(score = 12) = 2 P(1,11) P(3,9) P(5,7) P(2,10) P(4,8) P(6,6)     = 2 1 2 2 22 x x x 5 + x18 18 18 18 27 (c) P(score = 11|one of the die is blue) =P(score 11 one of the die is blue) P(one of the die is blue)   P(one of the die is blue)=P(RB) + P(WB) = 1 1 2x x 2 x 23 2 3  P(score 11 one of the die is blue)  = P(R1,B10) + P(R3,B8) + P(R5,B6) + P(W1,B10) + P(W3,B8) + P(W5,B6) + P(W7,B4) + P(W9, B2)

River Valley High School, Mathematics Department 2023 Probability T6 = 1 1 2x x 2 x 818 12 27  Required probability = 2 2 1 27 3 9 

River Valley High School, Mathematics Department 2023 Probability T7 9(i) P(C,C or N,N) + P(M,M or E,E)2 1 3 2 42 212 11 12 11 33                 (ii) P(C, C’, C) + P(C’, C, C) + P(C’, C’, C)2 10 1 10 9 2 12 12 11 10 12 11 10 6                  (iii) 3 2 3 3 12 11 10 220   (iv) 3 93!220 110  Let p = P(≤ n more cards are drawn to get an N) n p 1 2 10 < 0.75 2 2 8 2 17 10 10 9 45       < 0.75 3 17 8 7 2 8 45 10 9 8 15        < 0.75 4 8 8 7 6 2 2 5 10 9 8 7 3         < 0.75 5 2 8 7 6 5 2 7 3 10 9 8 7 6 9          > 0.75 Hence, the least value of n = 5 10(a) A and B are mutually exclusive. A and C are indept, i.e. P(A C) P(A) P(C)  1 1 7P(A) , P(B) , P(A C)5 10 15    23P(B C) 60  1 1 3P(A ) 5 10 10B   

River Valley High School, Mathematics Department 2023 Probability T8 P(A C) P(A) P(C) P(A C) 7 1 1P(C) P(C)15 5 5 4 5 1P(C) 15 4 3 1 1 1P(A C) x5 3 15                P(B C) P(B) P(C) P(B C) 23 1 1 P(B C)60 10 3 1P(B C) 20 1 1 1P(B) x P(C) x P(B C)10 3 30                Therefore, B and C are not independent. (b)(i) P(1st vase is flawless)1 2 1 1 7 2 3 2 2 12                   (ii) P(batch from X | 1st vase is flawless) =P(batch from X 1st vase is flawless) P(1st vase is flawless)  1 2 7 4 2 3 12 7          (iii) P(2 nd vase is flawless | 1st vase is flawless) =P(both vases are flawless) P(1st vase is flawless) 2 21 2 1 1 7 25 2 3 2 2 12 42                        11(i) P(2 balls are white) = 2 5 1 2