PROB NON CALC ANS SL

Transcription

1 IB Questionbank Maths SL PROB NON CALC ANS SL 0 min 0 marks. (a) P(X = ) = = A N 7 (b) P(X = ) = P(X = k) = k setting the sum of probabilities = M k e.g. + + =, + k = k k 9 = 9 accept = A k = AG N0 ( ) = (c) correct substitution into E X = xp( X x) A e.g = 7 ( ) = E X A N [7]

2 . (a) (i) s = A N (ii) evidence of appropriate approach e.g., + 8 q = q = A N (iii) p = 7, r = AA N (b) (i) P(art music) = A N 8 (ii) METHOD P = ( art) = evidence of correct reasoning e.g. 8 A R the events are not independent AG N0 METHOD P(art) P(music) = 9 = 8 A evidence of correct reasoning e.g. 8 R the events are not independent AG N0 (c) P(first takes only music) = = (seen anywhere) A 7 P(second takes only art)= (seen anywhere) A evidence of valid approach 7 e.g. 7 P(music and art)= = A N 0 80 []

3 . (a) (i) n 0. A N (ii) m 0., p 0., q 0. AAA N (b) appropriate approach e.g. P(B ) = P(B), m + q, (n + p) P(B ) = 0. A N. (a) (i) p = 0. A N (ii) q = 0. A N (iii) r = 0. A N (b) P(A B ) = A N 0. Note: Award A for an unfinished answer such as. 0. (c) valid reason R e.g. 0., thus, A and B are not independent AG N0 7. (a) (i) A N (ii) evidence of multiplying along the branches 7 e.g., 8 8 adding probabilities of two mutually exclusive paths 7 e.g. +, P(F) = A N

4 (b) (i) 8 A (ii) recognizing this is P(E F) 7 e.g. 8 7 = A N (c) X (cost in euros) 0 P (X) AA N (d) correct substitution into E(X) formula e.g , E(X) = (euros) A N []. (a) p = A N (b) multiplying along the branches e.g., 0 adding products of probabilities of two mutually exclusive paths e.g. +, P(B) = = A N 0 0

5 (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) = P(A B) = = 7 A N [7] 7. (a) P(A) = A N (b) P(B A) = A N 0 (c) recognising that P(A B) = P(A) P(B A) correct values e.g. P(A B) = 0 P(A B) = 0 A N 8. (a), 9, 9, 9, 0, 0, 0, 0, 0, 0 A N (b),,, (accept,,,,,,,, ) A N (c) P() =, P() =, P() =, P() = A N

6 (d) correct substitution into formula for E(X) A e.g. E(S) = E(S) = A N 9 (e) METHOD correct expression for expected gain E(A) for game e.g E(A) = 9 amount at end = expected gain for game = 00 (dollars) A N METHOD attempt to find expected number of wins and losses e.g., 9 attempt to find expected gain E(G) e.g E(G) = 00 (dollars) A N [] 9. (a) appropriate approach e.g. tree diagram or a table P(win) = P(H W) + P(A W)) = (0.)(0.8) + (0.)(0.) A = 0.0 (or 0.) A N

7 (b) evidence of using complement e.g. p, 0.9 choosing a formula for conditional probability P( W H ) e.g. P(H W ) = P( W ) correct substitution e.g. (0.)(0.7) 0. 0 = P(home) = 0.99 A N A [8] 0. (a) Note: Award A for vertical line to right of mean, A for shading to right of their vertical line. AA N (b) evidence of recognizing symmetry e.g. 0 is one standard deviation above the mean so d is one standard deviation below the mean, shading the corresponding part, 0 00 = 00 d d = 9 A N (c) evidence of using complement e.g. 0., p P(d < X < 0) = 0.8 A N. (a) (i) evidence of substituting into n(a B) = n(a) + n(b) n(a B) e.g , Venn diagram 0 A N (ii) A N

8 (b) (i) METHOD evidence of using complement, Venn diagram e.g. p, = 00 0 A N METHOD attempt to find P(only one sport), Venn diagram e.g = 0 A N 9 (ii) = A N 70 (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 A 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 AA N e.g., []. (a) (i) P(B) = A N (ii) P(R) = A N

9 (b) p = A N s =, t = A N (c) (i) P(X = ) = P (getting and ) = A = AG N0 (ii) P(X = ) = + or = A N (d) (i) X P(X = x) A N (ii) evidence of using E(X) = xp(x = x) E(X) = + = = A N

10 (e) win $0 scores one time, other time P() P() = (seen anywhere) A evidence of recognizing there are different ways of winning $0 e.g. P() P() + P() P(),, P(win $0) = = A N 8. (a) (i) correct calculation e.g , P(male or tennis) = = A N 0 (ii) correct calculation e.g. +, 0 0 P(not football female) = A N (b) 0 P(first not football) =, P(second not football) = 0 9 A 0 P(neither football) = 0 9 A 0 P(neither football) = = A N 80 8 [7]. (a) evidence of using p i = correct substitution A e.g. 0k + k + 0. =, 0k + k 0. = 0 k = 0. A N

11 (b) evidence of using E(X) = p i x i correct substitution e.g E(X) =. A N [7]. (a) evidence of binomial distribution (seen anywhere) e.g. X ~ B, mean = ( = 0.7) A N (b) P(X = ) = 9 P(X = ) = 0. = A N (c) evidence of appropriate approach M e.g. complement, P(X = 0), adding probabilities P(X = 0) = (0.7) 7 = 0., 7 P(X ) = 0.78 = A N [7]. (a) P(A B) = P(A) P(B) (= 0.x) A N (b) (i) evidence of using P(A B) = P(A) + P(B) P(A)P(B) correct substitution A e.g = 0. + x 0.x, 0. = 0.x x = 0. A N (ii) P(A B) = 0. A N

12 (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 P(X = ) = A N (ii) number of ways of getting X > is 7 P(X > ) = = A N (iii) P(X =7 X > ) = = 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 A 0 e.g. + k = 0, + 0k = 0 k = (=.) A N 0 [] 8. METHOD (a) σ = 0. 0 =. A x =. A N (b) 00. = 88.8 A N

13 METHOD (a) σ = 0 Evidence of using standardisation formula x 00 =. 0 A x =. A N 00 x (b) =. A 0 x = 88.8 A N 9. (a) For summing to e.g x = 0 x = A N 0 (b) For evidence of using E(X) = x f (x) Correct calculation A e.g E(X) = ( =.) A N 0 (c) A N [7] 0. (a) Evidence of using the complement e.g. 0.0 p = 0.9 A N (b) For evidence of using symmetry Distance from the mean is 7 e.g. diagram, D = mean 7 D = 0 A N

14 (c) P(7 < H < ) = = 0. A E(trees) = = 88 A N [9]. (a) (i) Attempt to find P(H) = = A N 7 (ii) Attempt to find P(H, T) = A = A N 9 (b) (i) Evidence of using np expected number of heads = A N (ii) heads, so 8 tails E(winnings) = 0 8 (= 0 8) = $ 8 A N [0]. (a) A N (b) P(A B) = P(A) + P(B) P(A B) P(A B) = P(A) + P(B) P(A B) 7 = + A 8 = (0.7) A N 0

15 P( A B) (c) P(A B) = = 0 A P( B) = (0.7) A N 0. (a) ( = 0.7) AA N 97 (b) ( = 0.) AA N 9 (c) ( = 0.08) A N (a) ( = 0.8) A N 0 (b) ( )(= ) Probability = = = = 0. A N 0 8 (c) Number studying = 7 Number not studying = 0 number studying = Probability = = = A N

16 . (a) /9 A /0 A /9 B /0 /9 A B /9 B AAA N (b) MM 8 8 =,0. 90 A N. (a) For summing to eg 0. + a b = a + b = 0. A N (b) evidence of correctly using E( X ) = x f ( x) eg a b, 0. + a b =. Correct equation 0 + a b =. (a + b = 0.9) Solving simultaneously gives a = 0. b = 0. AA N 7. (a) Independent P(A B) = P(A) P(B) (= ) = 0. A N

17 (b) P(A B) = P(A) + P(B) P(A B) (= ) M = 0.8 A 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 8. (a) For using p = (0. + p = ) p = 0. A N (b) For using E(X) = xp( X = x) E(X) = (0.) + (0.) + (0.) + (0.07) + (0.0) = A N A 0 9. (a) P(P C) = A 0+ 0 = A N 0 (b) P(P C ) = A 0+ 0 = A N (c) Investigating conditions, or some relevant calculations P is independent of C, with valid reason A N eg P(P C) = P(P C ), P(P C) = P(P), 0 0 = (ie P(P C) = P(P) P(C))

18 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 = ) M k = 0. A N 0 + (c) (i) P( X = 0) = 0 = A N 0 (ii) Evidence of using P(X > 0) = P(X = 0) 0 or = A N 0 [8]. (a) R M N 0 G R 8 0 G AAA N

19 (b) (i) P(M and G) = ( = = 0.) A N 8 (ii) P(G) = = = = 0. 7 A N (iii) P( M G) P(M G) = = P( G) = or 0. A N (c) P(R) = = Evidence of using a correct formula M E(win) = + or A 0 = $ accept, A N []. (a) For attempting to use the formula (P(E F) = P(E)P(F)) Correct substitution or rearranging the formula P E F eg P(F), P(F) =, P(F) = P E = ( ) ( ) P(F) = A N A

20 (b) For attempting to use the formula (P(E F) = P(E) + P(F) (P(E F)) P(E F) = + A = ( = 0.8) A N. (a) (i) Attempt to set up sample space, Any correct representation with pairs A N eg,,,,,,,,,,,,,,,, (ii) Probability of two s is (= 0.0) A N (b) x 0 P(X = x) 9 AAA N

21 (c) Evidence of selecting appropriate formula for E(X) eg E(X) = x P = 0 ( X x), E(X) = np Correct substitution 9 eg E(X) = 0 + +, E(X) = 8 E(X) = = A N [0]. (a) Using E(X) = x P ( X = x) 0 Substituting correctly E(X) = A = 8 (0.8) A 0 (b) (i) R R G G R G Note: Award for each complementary pair of probabilities, ie and, and, and. AAA

22 (ii) P(Y = 0) = = A 0 P(Y = ) = P(RG) + P(GR) = + M = A 0 P(Y = ) = = 0 For forming a distribution M y 0 P(Y = y) (c) P(Bag A) = = P(BagA B) = = For summing P(A RR) and P(B RR) Substituting correctly P(RR) = A = 7, 0. A 90 0 (d) P( A RR) For recognising that P( or RR) = P(A RR) = P( RR) = A =, 0. A 7 9 [9]. Total number of possible outcomes = (may be seen anywhere) (a) P( E ) = P(,) + P(, ) + P(, ) + P(, ) + P(, ) + P(, ) = (C)

23 (b) P( F ) = P(, ) + P(, ) + P(, ) = (C) (c) ( ) P E F = P( E) + P( F) P( E F) P( E F) = P = + ( E F ) 8 = =, 0. 9 (C) (a) (i) P( A) = = = 0.8 (N) 0 (ii) P(year art) = = = (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) = = = OR P( A)=P( A B) P( A B ) = (to be independent)

24 OR P( B)=P( B A) (to be independent) 00 0 P( B) = = = 0.7, P( B A ) = Note: Award the first only for a mathematical interpretation of independence. (b) n (history) = P(year history) = = = (N) (c) = = = 99 ( 0.0) (N) [] 0 7. Correct probabilities,,, Multiplying 0 70 P( girls) = = = 0.07 (C) 0 9

25 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 x = 0. ( x =., not possible) (A) P(B) = x = 0. (C) 9. (a) (b) (i) P(R S) = = = 0.7 (N) (ii) P(S) = + = ( = 0.) (N) 0 (iii) P(R S) = 0 8 = ( = 0.) (N) [0]

26 0. (a) P(A B) = P(A) + P(B) P(A B) 7 P(A B) = + 8 = (C) 8 P( A B) (b) P(A B) = = 8 P( B) = (C) (c) Yes, the events are independent (C) EITHER P(A B) = P(A) (R) (C) OR P(A B) = P(A)P(B) (R) (C). (a) L W W' L' 7 Note: Award for the given probabilities,, in the 8 correct positions, and for each bold value. L' L

27 7 (b) Probability that Dumisani will be late is = (0.9) (N) 0 (c) P(W L) = P(W L) = P(L) = 7 0 P(W L) = P( W L) P( L) = (= 0.7) (N) 7 [] 0. (a) = = 0. (C) 0 (b) = = = (A) (C) (c) 90 = = Accept 7 (C)

28 . (a) 0.9 Grows 0. Red Does not grow Grows 0. Yellow 0. Does not grow (A) (N) (b) (i) (ii) = 0. (N) ( = ) = 0.8 (N) (iii) P(red grows) P(grows) (may be implied) 0. = 0.8 = (N) 7 [0]

29 . (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). (a) E() H(8) U(88) a b c 9 n (E H) = a + b + c = 88 9 = 9 n (E H) = + 8 b = = b = a = = c = 8 = 7 Note: Award (A) for correct answers with no working. (b) (i) P(E H) = = 88 8 P( H ' E) (ii) P(H E) = = 88 P( E) 88 = (= 0.) OR Required probability =

30 (c) (i) P(none in economics) = = 0. Notes: Award (M0)(A0)(ft) for = Award no marks for (ii) P(at least one) = 0. = 0.77 OR = 0.77 [] 7 7. P(RR) = = P(YY) = = P (same colour) = P(RR) + P(YY) = (= 0.70, sf) (C) 7 7 Note: Award C for + =. 7. (a) P = (= 0.97 ( sf)) (A) (C)

31 (b) R R G etc G OR P = P (RRG) + P (RGR) + P (GRR) = (= 0.0 ( sf)) (C) 00

32 8. Sample space ={(, ), (, )... (, ), (, )} (This may be indicated in other ways, for example, a grid or a tree diagram, partly or fully completed)..... (a) P (S < 8) = 7 = OR P (S < 8) = 7 (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) = P S < 8 ( ) 7 = 7 = [7]

33 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) 0. P(different colours) = [P(GG) + P(RR) + P(WW)] = = 0 = (= 0.77, to sf) (C) OR P(different colours) = P(GR) + P(RG) + P(GW) + P(WG) + P(RW) + P(WR) = + = (= 0.77, to sf) (C) []. (a) s = 7.( sf) (G) (b) Weight (W) W 8 W 90 W 9 W 00 W 0 W 0 W Number of packets

34 (c) (i) From the graph, the median is approximately 9.8. Answer: 97 (nearest gram). (A) (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 = 0 i ( Wi W ) = Wi (e) Let A be the event: W > 00, and B the event: 8 < W 0 P( A B) P(A B) = P( B) P(A B) = P(B) = P(A B) = 0.8 OR 7 packets with weight 8 < W 0. Of these, 0 packets have weight W > 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. []. (a) U A B (C)

35 (b) n(a B) = n(a) + n(b) n(a B) = n(a B) n(a B) = (may be on the diagram) n(b A ) = 0 = (C) n( B A ) (c) P(B A ) = = = 0. (C) n( U ) 00 []. (a) (C) (b) P(B) = 0.(0.) + 0. (0.) = = 0. (C) P( B C) 0. (c) P(C B) = = = (= 0., sf) (C) P( B) 0. 9 []. (a) Males Females Totals Unemployed Employed Totals Note: Award if at least entries are correct. Award (A) if all 8 entries are correct. 0 (b) (i) P(unemployed female) = = 00

36 90 9 (ii) P(male I employed person) = = 0 []. (a) Boy Girl Total TV 8 Sport 9 Total 00 P(TV) = 8 00 (C) (b) P(TV Boy) = (= 0.8 to sf) (A) (C) Notes: Award for numerator and for denominator. Accept equivalent answers. []. (a), not not, not not, not not, not Notes: Award for probabilities correctly entered on diagram. Award for correctly listing the outcomes, ; not ; not, ; not, not, or the corresponding probabilities., (M) (C)

37 (b) P(one or more sixes) = + + or = (C) [] 7. (a) A B (C) (b) (i) n(a B) = (C) (ii) P(A B) = or (allow ft from (b)(i)) (C) 8 (c) n(a B) 0 (or equivalent) (R) (C) [] 7 8. p(red) = = p(black) = = (a) (i) p(one black) = 8 8 = 0.9 to sf (ii) p(at least one black) = p(none) = = 0. = 0.

38 00 (b) 00 draws: expected number of blacks = 8 = 0 [8] 9. (a) p(a B) = = 0. (C) (b) p( A B) = p( (A B)) = 0. = 0. (C) []