. (a) P(X = ) 7 N (b) P(X = ) = k P(X = k) = setting the sum of probabilities = () () M e.g. k =, 5 + k = k k = 9 accept 9 k = AG N0 (c) correct substitution into E xp( X X x) e.g. 9 8 EX 7 N [7]. (a) (i) s = N (ii) evidence of appropriate approach e.g., + 8 q =5 q = 5 N (iii) p = 7, r = N 5 (b) (i) P(art music) = 8 5 A N IB Questionbank Maths SL
(ii) METHOD P art evidence of correct reasoning e.g. 5 8 the events are not independent AG N0 R METHOD P(art) P(music) = 9 5 8 evidence of correct reasoning e.g. 8 5 the events are not independent AG N0 R (c) P(first takes only music) = = (seen anywhere) 7 P(second takes only art)= (seen anywhere) 5 evidence of valid approach 7 e.g. 5 7 P(music and art)= 0 80 N []. (a) (i) n = 0. N (ii) m = 0., p = 0., q = 0. N (b) appropriate approach e.g. P(B ) = P(B), m + q, (n + p) P(B ) = 0. N. (a) (i) p = 0. N IB Questionbank Maths SL
(ii) q = 0. N (iii) r = 0. N (b) P(A B ) = A N Note: Award for an unfinished answer such as 0.. 0. (c) valid reason R e.g. 0.5, 0.5 0. thus, A and B are not independent AG N0 5. (a) (i) 7 N (ii) evidence of multiplying along the branches 5 7 e.g., 8 8 adding probabilities of two mutually exclusive paths 7 5 e.g., 8 8 8 8 P(F) = N (b) (i) 8 () IB Questionbank Maths SL
(ii) recognizing this is P(E F) 7 e.g. 8 7 A N (c) X (cost in euros) 0 P (X) 9 9 9 A N (d) correct substitution into E(X) formula e.g. 0, 9 9 9 9 9 E(X) = (euros) N []. (a) p = 5 N (b) multiplying along the branches e.g., 5 0 adding products of probabilities of two mutually exclusive paths e.g., 5 5 8 0 0 7 P(B) = 0 0 N IB Questionbank Maths SL
(c) appropriate approach which must include A (may be seen on diagram) P( A B) P( A B) e.g. do not accept P( B) P( B) P(A B) = 5 8 () 7 0 P(A B) = 7 N [7] 7. (a) P(A) = N (b) P(B A) = 0 A N (c) recognising that P(A B) = P(A) P(B A) correct values () e.g. P(A B) = 0 P(A B) = 0 N 8. (a), 9, 9 5, 9, 0, 0 5, 0, 0, 0 5, 0 A N (b),,, 5 (accept,,,,,,, 5, 5) A N (c) P() = 9, P() = 9, P() = 9, P(5) = 9 A N IB Questionbank Maths SL 5
(d) correct substitution into formula for E(X) e.g. E(S) = 5 9 9 9 9 E(S) = 9 A N (e) METHOD correct expression for expected gain E(A) for game 5 e.g. 50 0 9 9 50 E(A) = 9 () amount at end = expected gain for game = 00 (dollars) N METHOD attempt to find expected number of wins and losses 5 e.g., 5 9 attempt to find expected gain E(G) e.g. 50 0 0 E(G) = 00 (dollars) N [] 9. (a) appropriate approach e.g. tree diagram or a table P(win) = P(H W) + P(A W)) = (0.5)(0.8) + (0.5)(0.) = 0.05 (or 0.) N IB Questionbank Maths SL
(b) evidence of using complement e.g. p, 0.95 choosing a formula for conditional probability P( W H ) e.g. P(H W ) = P( W ) correct substitution (0.5)(0.7) 0. 05 e.g. 0.95 0. 95 P(home) = 0.99 N [8] 0. (a) Note: Award for vertical line to right of mean, for shading to right of their vertical line. N (b) evidence of recognizing symmetry e.g. 05 is one standard deviation above the mean so d is one standard deviation below the mean, shading the corresponding part, 05 00 = 00 d d = 95 N (c) evidence of using complement e.g. 0., p P(d < X < 05) = 0.8 N. (a) (i) evidence of substituting into n(a B) = n(a) + n(b) n(a B) e.g. 75 + 55 00, Venn diagram 0 N (ii) 5 N IB Questionbank Maths SL 7
(b) (i) METHOD evidence of using complement, Venn diagram e.g. p, 00 0 70 7 00 0 N METHOD attempt to find P(only one sport), Venn diagram 5 5 e.g. 00 00 70 7 00 0 N (ii) 5 9 70 A N (c) valid reason in words or symbols (R) e. g. P(A B) = 0 if mutually exclusive, P(A B) if not mutually exclusive correct statement in words or symbols N e.g. P(A B) = 0., P(A B) P(A) + P(B), P(A) + P(B) >, some students play both sports, sets intersect (d) valid reason for independence (R) e.g. P(A B) = P(A) P(B), P(B A) = P(B) correct substitution N 0 75 55 0 75 e.g., 00 00 00 55 00 []. (a) (i) P(B) = (ii) P(R) = N N IB Questionbank Maths SL 8
(b) p N s, t N (c) (i) P(X = ) = P (getting and ) = = AG N0 (ii) P(X = ) = + = or () N (d) (i) X P(X = x) A N (ii) evidence of using E(X) = xp(x = x) E(X) = () 5 = N IB Questionbank Maths SL 9
(e) win $0 scores one time, other time P() P() = (seen anywhere) evidence of recognizing there are different ways of winning $0 e.g. P() P() + P() P(),, 5 5 5 5 P(win $0) = 78 5 9 8 N. (a) (i) correct calculation () e.g. 9 5, 0 0 0 0 P(male or tennis) = 0 5 (ii) correct calculation () e.g., 0 0 P(not football female) = N N 0 (b) P(first not football) =, P(second not football) = 0 9 0 P(neither football) = 0 9 P(neither football) = 0 80 8 N [7]. (a) evidence of using p i = correct substitution e.g. 0k + k + 0. =, 0k + k 0. = 0 k = 0. A N (b) evidence of using E(X) = p i x i correct substitution () IB Questionbank Maths SL 0
e.g. 0. + 0. + 0. E(X) =.5 N [7] 5. (a) evidence of binomial distribution (seen anywhere) e.g. X ~ B, N mean = 0.75 (b) P(X = ) = () P(X = ) = 0. 9 N (c) evidence of appropriate approach M e.g. complement, P(X = 0), adding probabilities P(X = 0) = (0.75) 7 0., () P(X ) = 0.578 7 N [7]. (a) P(A B) = P(A) P(B) (= 0.x) N (b) (i) evidence of using P(A B) = P(A) + P(B) P(A)P(B) correct substitution e.g. 0.80 = 0. + x 0.x, 0. = 0.x x = 0.5 N (ii) P(A B) = 0. N IB Questionbank Maths SL
(c) valid reason, with reference to P(A B) R N e.g. P(A B) 0 7. (a) (i) number of ways of getting X = is 5 () 5 P(X = ) = N (ii) number of ways of getting X > is () 7 P(X > ) = N (iii) P(X =7 X > 5) = A N (b) evidence of substituting into E(X) formula 0 finding P(X < ) = (seen anywhere) (A) evidence of using E(W) = 0 correct substitution 5 0 e.g. k 0, 5 + 0k = 0 A k = (=.) N 0 [] 8. METHOD (a) σ = 0 (). 0 =.. + 00 x =. N (b) 00. = 88.8 N IB Questionbank Maths SL
METHOD (a) σ = 0 () Evidence of using standardisation formula x 00. 0 x =. N (b) 00 x =. 0 x = 88.8 N 9. (a) For summing to e.g. x 5 5 0 x = 0 N (b) For evidence of using E(X) = x f (x) Correct calculation e.g. 5 5 0 0 E(X) = (.5) 0 N (c) 0 0 00 N [7] 0. (a) Evidence of using the complement e.g. 0.0 p = 0.9 N (b) For evidence of using symmetry Distance from the mean is 7 () e.g. diagram, D = mean 7 D = 0 N IB Questionbank Maths SL
(c) P(7 < H < ) = 0.5 0.0 = 0. E(trees) = 00 0. = 88 N [9]. (a) (i) Attempt to find P(H) = = 7 N (ii) Attempt to find P(H, T) = = 9 N (b) (i) Evidence of using np expected number of heads = N (ii) heads, so 8 tails () E(winnings) = 0 8 (= 0 8) = $ 8 N [0]. (a) N (b) P(A B) = P(A) + P(B) P(A B) P(A B) = P(A) + P(B) P(A B) = 7 5 8 = 0 (0.75) N (c) P(A B) = P( A B) P( B) 0 IB Questionbank Maths SL
= (0.7) 0 N N 97. (a) 0.7 N 5 (b) 0.55 59 A N 97 (c) 0.08 9 N 0. (a) 0.58 (b) 5 (8 + 5 + 7)(= 5) Probability = 5 0 0. 5 8 N (c) Number studying = 7 () Number not studying = 0 number studying = Probability = 0 0. 7 0 N IB Questionbank Maths SL 5
5. (a) /9 A /0 A /9 B /0 /9 A B 5/9 B N (b) 0 9 0 9 MM 8 8,0. 5 90 5 N -. (a) For summing to eg 0. + a + 0. + b = a + b = 0. (b) evidence of correctly using E( X ) x f ( x) eg 0 0. + a + 0. + b, 0. + a + 0. + b =.5 Correct equation 0 + a + 0. + b =.5 (a + b = 0.9) Solving simultaneously gives a = 0.5 b = 0.5 IB Questionbank Maths SL
50 t < 0 () (Total marks) 7. (a) Independent P(A B) = P(A) P(B) (= 0. 0.8) = 0. N (b) P(A B) = P(A) + P(B) P(A B) (= 0. + 0.8 0.) M = 0.8 N (c) No, with valid reason A N eg P(A B) 0 or P(A B) P(A) + P(B) or correct numerical equivalent IB Questionbank Maths SL 7
8. (a) For using p (0. + p + 0. + 0.07 + 0.0 = ) (b) For using E(X) = X x p = 0. N xp E(X) = (0.) + (0.) + (0.) + (0.07) + 5(0.0) = A N 9. (a) P(PC) = 0 0 0 = N (b) P(PC) = 0 00 = N (c) Investigating conditions, or some relevant calculations P is independent of C, with valid reason N eg P(PC) = P(PC), P(PC) = P(P), 0 50 0 (ie P(P C) = P(P) P(C)) 50 50 50 0. (a) Adding probabilities Evidence of knowing that sum = for probability distribution R eg Sum greater than, sum =., sum does not equal N (b) Equating sum to (k + 0.7 = ) M k = 0. N IB Questionbank Maths SL 8
(c) (i) P 0 0 X 0 = 0 (ii) Evidence of using P(X > 0) = P(X = 0) 5 0 or 0 0 0 9 = 0 N N [8]. (a) 5 R M N 5 0 G R 8 0 G N IB Questionbank Maths SL 9
(b) (i) P(M and G) = ( 0.) N 5 5 (ii) P(G) = (iii) P(M G) = 8 ()() 5 0 0 = 0. 7 5 P( M G) P( G) 5 ()() = 5 or 0. N N (c) P(R) = () Evidence of using a correct formula M 8 E(win) = 5 or 5 5 5 0 5 0 = $ 0 accept, 5 N []. (a) For attempting to use the formula (P(E F) = P(E)P(F)) Correct substitution or rearranging the formula eg P(F), P(F) = E F PE P(F) = P, P(F) = N IB Questionbank Maths SL 0
(b) For attempting to use the formula (P(E F) = P(E) + P(F) (P(E F)) P(E F) = 5 N = 0.8. (a) (i) Attempt to set up sample space, Any correct representation with pairs A N eg,,,,,,,,,,,,,,,, (ii) Probability of two s is (= 0.05) N (b) x 0 P(X = x) 9 N IB Questionbank Maths SL
(c) Evidence of selecting appropriate formula for E(X) eg E(X) = PX x Correct substitution 0 x, E(X) = np 9 eg E(X) = 0 + +, E(X) = E(X) = 8 N [0]. (a) Using E(X) = x P( X x) 0 Substituting correctly E(X) = 0 0 0 0 = 8 (0.8) 0 (b) (i) 5 R R 5 G G 5 R 5 G Note: Award () for each complementary pair of probabilities, ie and, and, and. 5 5 5 5 IB Questionbank Maths SL
(ii) P(Y = 0) = 5 5 0 P(Y = ) = P(RG) + P(GR) = 0 P(Y = ) = M 5 5 () 5 0 For forming a distribution M 5 y 0 P(Y = y) 0 0 0 (c) P(Bag A) = () P(BagA B) = For summing P(A RR) and P(B RR) Substituting correctly P(RR) = () 0 0 = 7, 0. 90 0 5 (d) For recognising that P( or RR) = P(A RR) = P( A RR) P( RR) = 7 0 90 =, 0. 7 9 [9] 5. Total number of possible outcomes = (may be seen anywhere) () (a) P( E) P(,) P(, ) P(, ) P(, ) P(5, 5) P(, ) () (C) (b) P( F) P(, ) P(5, 5) P(, ) () (C) IB Questionbank Maths SL
(c) P E F P( E) P( F) P( E F) P( EF) () 8 PE F, 0. 9 () (C). (a) (i) 80 8 P( A) 0.8 0 () (N) (ii) 5 P(year art) 0.7 0 () (N) (iii) No (the events are not independent, or, they are dependent) () (N) EITHER P( A B) P( A) P( B) (to be independent) 00 0 P( B) 0.7 0 () 8 0 () OR P( A)=P( A B ) (to be independent) 5 P( AB) () 00 8 5 () 00 IB Questionbank Maths SL
OR P( B)=P( B A ) (to be independent) 00 0 P( B) 0.7 0, 5 P( B A) () 80 5 00 80 0 () Note: Award the first only for a mathematical interpretation of independence. (b) n(history) 85 () 50 0 P(year history) 0.588 85 7 ()(N) (c) 0 00 00 0 0 00 0 09 0 09 0 09 ()() 00 0.50 99 () (N) [] 0 7. Correct probabilities,,, 0 Multiplying 70 5 P( girls) = 0.07 550 9 ()()()() () (C) IB Questionbank Maths SL 5
8. For using P(A B) = P(A) + P(B) P(A B) Let P(A) = x then P(B) = x P(A B) = P(A) P(A) (= x ) () 0.8 = x + x x () x x + 0.8 = 0 x = 0. ( x =., not possible) (A) P(B) = x = 0. () (C) 9. (a) ()()() (b) (i) P(R S) = 0.7 5 5 () (N) (ii) P(S) = 5 ()() (iii) P(R S) = = ( 0.) () (N) 0 5 0 ()() 8 = ( 0.5) () (N) [0] 0. (a) P(A B) = P(A) + P(B) P(A B) IB Questionbank Maths SL
P(A B) = = 8 7 8 () (C) (b) P(AB) = P( A B) P( B) 8 = () (C) (c) Yes, the events are independent () (C) EITHER P(AB) = P(A) (R) (C) OR P(A B) = P(A)P(B) (R) (C). (a) L 7 8 8 W W' 5 5 L' ()()()() 7 Note: Award () for the given probabilities,, 8 5 in the correct positions, and () for each bold value. L' L IB Questionbank Maths SL 7
(b) Probability that Dumisani will be late is 7 ()() 8 8 5 7 = (0.9) 0 () (N) (c) P(WL) = P(W L) = 7 P(L) = 0 P( W L) P( L) 7 () 8 () P(WL) = 7 7 0 5 = (= 0.75) () (N) 7 []. (a) (b) 0 0. 0 90 0 0 7 0.58 0 0 ()() (C) (A) (C) (c) 90 0.9 0 7 Accept 7 ()() (C) IB Questionbank Maths SL 8
. (a) 0.9 Grows 0. Red 0. 0.8 Does not grow Grows 0. Yellow 0. Does not grow (A) (N) (b) (i) 0. 0.9 () = 0. () (N) (ii) 0. 0.0.8 ( 0. 0.8) () (iii) = 0.8 () (N) P(red grows) P(grows) (may be implied) 0. () 0.8 0.9 7 () (N) 7 [0] IB Questionbank Maths SL 9
. (a) Independent (I) (C) (b) Mutually exclusive (M) (C) (c) Neither (N) (C) Note: Award part marks if the candidate shows understanding of I and/or M eg I P(A B) = P(A)P(B) M P(A B) = P(A) + P(B) 5. (a) E() H(8) U(88) a b c 9 n (E H) = a + b + c = 88 9 = 9 n (E H) = + 8 b = 9 0 9 = b = () a = = () c = 8 = 7 () Note: Award (A) for correct answers with no working. (b) (i) P(E H) = (ii) P(HE) = () 88 8 P H ' P E 88 88 (= 0.5) () E = OR Required probability = ()() IB Questionbank Maths SL 0
(c) (i) 5 55 5 P(none in economics) = 8887 8 () = 0.5 () 5 Notes: Award (M0)(A0)()(ft) for 88 5 55 5 Award no marks for. 888888 = 0.58. (ii) P(at least one) = 0.5 = 0.77 () OR 5 55 5 0 88 87 8 88 87 8 88 87 8 = 0.77 () 5 [] 7 7. P(RR) = 5 5 P(YY) = P (same colour) = P(RR) + P(YY) () () = (= 0.70, sf) () (C) Note: Award C for 7 5 7. 7. (a) P = (= 0.957 ( sf)) (A) (C) IB Questionbank Maths SL
(b) R 5 R G 5 OR P = P (RRG) + P (RGR) + P (GRR) G etc () 5 5 5 9 = (= 0.0 ( sf)) () (C) 00 IB Questionbank Maths SL
8. Sample space ={(, ), (, )... (, 5), (, )} (This may be indicated in other ways, for example, a grid or a tree diagram, partly or fully completed) 5..... 5 (a) P (S < 8) = 7 = OR 7 P (S < 8) = () (A) (b) P (at least one ) = = OR P (at least one ) = () (A) P(at least one S 8) (c) P (at least one S < 8) = PS 8 7 = 7 = () () [7] IB Questionbank Maths SL
9. (a) P (A B) = P (A) + P (B) P (A B) P (A B) = P (A) + P (B) P (A B) = = (0.0909) () (C) (b) For independent events, P (A B) = P (A) P (B) = () = (0.099) () (C) 50. P(different colours) = [P(GG) + P(RR) + P(WW)] 0 9 0 9 5 = 5 5 5 () 0 = 50 () = (= 0.77, to sf) 5 () (C) OR P(different colours) = P(GR) + P(RG) + P(GW) + P(WG) + P(RW) + P(WR) () 0 0 0 = ()() 5 5 = (= 0.77, to sf) () (C) 5 [] 5. (a) s = 7.( sf) (G) (b) Weight (W) W 85 W 90 W 95 W 00 W 05 W 0 W 5 Number of packets 5 5 0 5 9 7 80 () (c) (i) From the graph, the median is approximately 9.8. Answer: 97 (nearest gram). (A) IB Questionbank Maths SL
(ii) From the graph, the upper or third quartile is approximately 0.. Answer: 0 (nearest gram). (A) (d) Sum = 0, since the sum of the deviations from the mean is zero. (A) OR W i ( W i W ) Wi 80 = 0 80 () (e) Let A be the event: W > 00, and B the event: 85 < W 0 P( A B) P(AB) = P( B) 0 P(A B) = 80 7 P(B) = 80 P(AB) = 0.8 () () () OR 7 packets with weight 85 < W 0. Of these, 0 packets have weight W > 00. 0 Required probability = 7 () = 0.8 () Notes: Award (A) for a correct final answer with no reasoning. Award up to (M) for correct reasoning or method. [] 5. (a) U A B () (C) (b) n(a B) = n(a) + n(b) n(a B) 5 = 0 + 50 n(a B) n(a B) = 5 (may be on the diagram) n(b A) = 50 5 = 5 () (C) IB Questionbank Maths SL 5
(c) P(B A) = n( B A) n( U) 5 00 = 0.5 () (C) [] 5. (a) 0. 0. 0. 0. 0.5 0.5 () (C) (b) P(B) = 0.(0.) + 0. (0.5) = 0. + 0.0 = 0.5 () (C) (c) P(CB) = P( B C) P( B) 0. 0.5 9 (= 0., sf) () (C) [] 5. (a) Males Females Totals Unemployed 0 0 0 Employed 90 50 0 Totals 0 90 00 Note: Award () if at least entries are correct. Award (A) if all 8 entries are correct. (b) (i) P(unemployed female) = 0 () 00 5 (ii) P(male I employed person) = 90 9 () 0 [] 55. (a) Boy Girl Total IB Questionbank Maths SL
TV 5 8 Sport 9 Total 5 00 8 P(TV) = 00 () (C) (b) P(TV Boy) = (= 0.8 to sf) (A) (C) Notes: Award () for numerator and () for denominator. Accept equivalent answers. [] 5. (a) 5 not 5 5 Notes: Award for probabilities not not, 5,, not not, not, not correctly entered on diagram. Award for correctly listing the outcomes, ; not ; not, ; not, not, or the corresponding probabilities. 5 5 5 (M) (C) (b) 5 5 5 5 P(one or more sixes) = or = () (C) [] 57. (a) IB Questionbank Maths SL 7
A B () (C) (b) (i) n(a B) = () (C) (ii) P(A B) = or 8 (allow ft from (b)(i)) () (C) (c) n(a B) 0 (or equivalent) (R) (C) [] 58. p(red) = 5 7 5 p(black) = 0 8 0 8 7 8 7 (a) (i) p(one black) = () 8 8 = 0.9 to sf () (ii) p(at least one black) = p(none) 0 8 8 7 = 0 8 8 () = 0. = 0.5 () IB Questionbank Maths SL 8
00 (b) 00 draws: expected number of blacks = 8 = 50 () [8] 59. (a) p(a B) = 0. + 0.8 = 0. () (C) (b) p( A B) = p( (A B)) = 0. = 0. () (C) [] IB Questionbank Maths SL 9