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 (