TOPICAL REVISION PACKAGE III (SUGGESTED SOLUTIONS)

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1 Topical Revision Package III Suggested Solutions H Mathematics (7) PREPARATORY COURSE FOR SINGAPORE-CAMBRIDGE GENERAL CERTIFICATE OF EDUCATION (ADVANCED LEVEL) H MATHEMATICS (7) TOPICAL REVISION PACKAGE III (SUGGESTED SOLUTIONS) Version Page of

2 Topical Revision Package III Suggested Solutions H Mathematics (7) Permutations and Combinations ) JJC/II/ (i) ENDANGERED: E, N, D, A, G, R No. of -letter code-words from E, N, D, A, G, R C! 0 (ii) Select letter from N, D, A, G,R No. of -letter code-words with the chosen letter and E s (iii) Case I: same letters No. of such code-words = 0 Case II: pair of same letters! No. of code-words C C 0! Case III: pairs of same letters No. of code-words!!! C!! C 0 No. of -letter code-words that contain at least repeated letter = = ) MI/II/ (a) (i) Consider the green, yellow and purple tiles as unit. Number of ways to arrange the green, yellow and purple tiles within the unit!! Arranging the unit with the rest of the tiles Thus, no. of possible arrangement for the tiles 7!!!! 7! 00!!! Page of

3 Topical Revision Package III Suggested Solutions H Mathematics (7) (ii) Number of ways to arrange the red, green, yellow and purple tile Number of ways to slot in the blue tiles C!!! Number of possible arrangements such that no blue tiles are placed next to! another C 00!! (iii) Number of possible arrangements such that a red tile at the beginning and! another red tile at the end of the line 00.!! (b) (i) Number of ways = ( )! = 0 (ii) Insertion method Number of ways 7 7! C!! 00 (iii) Number of ways = ( )!! = 000 ) SRJC/II/ (a) Number of ways = = (b) (i) Number of parking arrangements (ii) =! (cases where the two cars are taking the side with lots) (cases where the two cars are taking the side with lots! C!!! C!!! Number of parking arrangements C!7! 00 ) DHS/II/ (i) Number of ways =! = 00 (ii) Number of ways to arrange the boys and girls on each side = Number of ways the group of boys and girls can arrange themselves =!! Required number of ways =!! = 0 (iii) Number of ways to choose any out of the seats to sit the boys and girls Number of ways to arrange the boys and girls! Required number of ways C! 00 C Page of

4 Topical Revision Package III Suggested Solutions H Mathematics (7) (iv) Number of ways to choose any out of the boys to sit on the seats and number of ways to arrange the boys C! Number of ways to choose any out of the seats to sit the remaining boys/ girls and number of ways to arrange the boys/ girls C! Required number of ways C! C! ) PJC/II/ Case : No. of code-words Case : No. of code-words Case : No. of code-words all letters are different C! 70 pair of repeated letters are used 7! C C 00! pairs of repeated letters are used! C C 0!! Total number of -letter code-words = = 0 ) RI/II/ (i) Number of ways for all man to stand next to his wife =! Required number of ways =!! = 70 (ii) Number of ways to choose the couple who don t stand next to each other C Number of ways to arrange the couples who are standing next to their spouse!! Number of ways to arrange each couple standing together =! = Number of ways to slot the remaining couple in between any two person P Required number of ways C! P Page of

5 Topical Revision Package III Suggested Solutions H Mathematics (7) 7) NYJC/II/ Number of ways = 7! + 7!! = 0 (i) Number of ways (ii) Number of ways!! 00!! 00! ) TJC/II/7 (a) (i) Number of ways C C C 0 (ii) There will be = empty seats Therefore, the number of ways of rearranging the three groups where each family of three sit together is!!!!! (b) Number of ways!! P 0 ) RVHS/II/7 (a) COORDINATION: O, I, N, A, C, R, D, T There are only possible arrangement format: (i) Consonant, Vowel, Consonant, Vowel,, Consonant, Vowel, Consonant, Vowel. (ii) Vowel, Consonant, Vowel, Consonant,, Vowel, Consonant, Vowel, Consonant. As there are Ns, O and Is, the total number of different arrangements in which the!! vowels and consonants alternates is 00.!!! (b) Case : consonants and vowels No. of possible passwords C C Case : consonants and vowel No. of possible passwords! 00 C C! Total no. of passwords with more than consonants = 00 + =. Page of

6 Topical Revision Package III Suggested Solutions H Mathematics (7) ) AJC/II/ (i) Family, Holiday, Friends 7 Choose images at least images from the Friends folder and at least image from each of the other two folders. Case : Friends + Holiday + Family No. of selections 7 C C C 0 Case : Friends + Holiday + Family No. of selections 7 C C C 0 Case : Friends + Holiday + Family No. of selections 7 C C C 0 Total no. of selections = = 70 (ii) Case : identical images in first row No. of arrangements =! Case : identical images in second row of No. of arrangements C! 0 Total no. of arrangements = = 00 (iii) Pimage A will appear a second time in the next n images 0. n 0. n 0. n 0. ln 0. n.7 ln The least value of n is. Page of

7 Topical Revision Package III Suggested Solutions H Mathematics (7) Probability ) CJC/II/7 (i) Pchosen serviceman is an Officer 0.0 (ii) 0.7[(0.0)(0.) + (0.)(0.7) + (0.7)(0.)] + (0.)(a)(0.) = a = 0.0 a = 0.0 Probability required P Failed NPFA an Officer P an Officer P Failed NPFA and an Officer P an Officer ( s.f.) 0.0 (iii) Probability required = = (iv) Probability Required P Failed NPFA a Non-Officer P a Non-Officer Pan Officer P Failed NPFA and an Officer ( s.f.) ( s.f.) ) SRJC/II/7 (a) (i) P A P A B P A B' P A P A B, A and B are independent events. AB P B Since (ii) P P B PB P A B P A PB P A B 7 Page 7 of

8 Topical Revision Package III Suggested Solutions H Mathematics (7) (b) (c) Required probability = (0.)(0.) + 0.7(0.)(0.7)(0.(0.) + Required probability = 0.[ + (0.7 0.) + (0.7 0.) + ] ! 7! C!! ) JJC/II/7 (i) For a set match, Probability of A winning = probability of B winning p(0.)(0.) + ( p)(0.)(0.7) = p(0.)(0.) + ( p)(0.)(0.) 0.p p = 0.p p 0.p = 0. p (Shown) (ii) P B wins the first set A wins the match P B wins the first set Awins the match P A wins the match = ( s.f.) Page of

Topical Revision Package III Suggested Solutions H Mathematics (7) ) IJC/II/ (i) P(the result is positive) = 0.7(0.) + 0.7(0.0) = 0. 0. ( s.f.

9 Topical Revision Package III Suggested Solutions H Mathematics (7) ) IJC/II/ (i) P(the result is positive) = 0.7(0.) + 0.7(0.0) = ( s.f.) (ii) P(elderly has diabetes given that the result is positive) P elderly has diabetes and the result is positive P the result is positive ( s.f.) ) PJC/II/7 Let event A be watched Jogging man Let event B be watched Voice of me Given: P(A) = 0.7, P(B) = 0., P(A B ) = 0. P A B' (i) P AB' 0. P B' P A B' A B A A B (ii) P A' B P A B' PB P A B P A B' P P P ' = (0. 0.) + 0. = 0. P B A' 0.0 P BA' 0. P A' 0. (iii) Since P(A) = 0.7 P(A B ) = 0., the two events are not independent. Page of

10 Topical Revision Package III Suggested Solutions H Mathematics (7) ) ACJC/II/ (i) Required probability n n n n n n nn (ii) Required probability n n n nn n 7) NJC/II/ (a) (b) (c) Required probability P(one gets the flu) = 0.p + 0.(0.) C C ( s.f.) 0 C = 0.(0.) + (0.)(0.) = 0. or 0 ( s.f.) P(exactly one of the two people took the flu vaccine both of them get the flu) ! 0. = ( s.f.) ) RI/II/ Let A and B be the event that a student takes Mathematics A and Mathematics B respectively. Let F be the event that a student failed the paper that he/she sat for. Page of

11 Topical Revision Package III Suggested Solutions H Mathematics (7) 0.p 0 p 0 0.p P p 0 0.p P F ' 0 0 P(exactly one out of failed) = P(one failed and one passed) 0 0.p 0 0.p C p p 0.0p 000 = 0.000(00 p 0.0p ) (Shown) P(both take Mathematics A exactly one out of failed) (i) F (ii) P one takes Mathematics Aand failed and the other takes Mathematics Aand passed P exactly one out of failed 0.p 0.7 p C p0.0 p 0.p 7 00 p0.0p Hence,.0p = 00 p 0.07p.p + p 00 = 0 From the G.C., p = 0 or 0 (rej since p 0) ) AJC/II/7 AB P A B' P B' (i) P ' P A B' PB Thus, P A B P A P A B' 0 Page of

12 Topical Revision Package III Suggested Solutions H Mathematics (7) (ii) P A' AB P P A' B A B PB P AB A B A B P P P 0 0 (iii) A is independent of C P A' C P A' PC (iv) P(A B C) is greatest when A C is totally in B. P A ' C P A' B C P(A B C) is least when A C is furthest away from B. Since P A' B' P A' B C 0. (Shown) 0 P A' B C ) DHS/II/7 (i) There are! ways to arrange the girls. There are possible spaces where boys can be slotted in. Choose spaces out of. There are! ways to arrange the boys. P all boys are separated! C!! or 0. ( s.f.) Page of

13 Topical Revision Package III Suggested Solutions H Mathematics (7) (ii) Choose out of girls to be on both sides of the particular boy. There are! ways to arrange the girls. Taking GBG as one group, there are! ways to arrange the group and other students. P a particular boy is between girls C!!! or 0. ( s.f.) (iii) Taking the boys as one group, there are! ways to arrange the boys. Taking the boys group as reference point, there are! ways to arrange the rest of the girls. P all boys next to one another in a circle (iv) Pat least student from each race!!! P Chinese and Malaly only or Chinese and Indian only C C 7 7 C7 or 0.0 ( s.f.) or 0. ( s.f.) Page of

14 Topical Revision Package III Suggested Solutions H Mathematics (7) Discrete Random Variable P p0 p p --- (). Since X x all w Given that E X 0 p p p 0p0 p p --- () 0 Given that Var X E X 0 p0 p p p p p --- () 0 From the G.C., solving (), () and (), we get p0, p and 0 p. The probability distribution table for R is r 0 P(R = r) ER r PR r 0 all r Var R E R 0 0 Page of

15 Topical Revision Package III Suggested Solutions H Mathematics (7). P (a) Since X x all x (Shown) E 0 (b) X or. 0 E X X X X Var E E..7 (c) The possible pairs are (, ), (,), (,), (, ). Required probability 7 or 0. ( s.f.). Note: PY 0 PY PY Hence, the probability distribution of Y is Y 0 P(Y = y) Then, EY y PY y 0 all y (Shown) Note: ET EY y PY y 0 all y Var Var E 0 T Y Y (Shown) Page of

16 Topical Revision Package III Suggested Solutions H Mathematics (7). (i) The table of outcome is Disc Die Note that the probabilty to get any value in the table of outcome is Hence, the probability distribution table of X is. (ii) P X X 7 r 7 P(R = r) P X X 7 P X 7 X X X X P X P P P P 7 (Shown) (iii) E X (since the distribution is symmetrical about X = ) Var X E X 7 (Shown) (iv) Var X X Var X Var X Var X (v) P XX P X 7P X P X P X P X P X Page of

17 Topical Revision Package III Suggested Solutions H Mathematics (7) P X x p + q =. (a) Note: (b) all x p + q = () Given E(X) =. (0.) + (0.) + p + q =. p + q = () From the G.C., solving () and (), we get p = 0. and q = 0.0. Var(X) = E[(X μ) ] all x x P X x = (.) (0.) + (.) (0.) + (.) (0.) + (.) (0.0) =. Let Y be the amount of money received by a player in on roll. The probability distribution of Y is y 0 P(Y = y) Note: EY Let n be the number of rolls a player has. We want E(Y + Y + + Y n ) = E(nY) = ne(y) = n n The player should be allowed rolls to receive $ Page 7 of

18 Topical Revision Package III Suggested Solutions H Mathematics (7) 7. (a) Red Ball drawn Non-Red Ball drawn Red Ball drawn Non-Red Ball drawn Red Ball drawn Non-Red Ball drawn Red Ball drawn Let R be the amount of points awarded. Hence, from the probability tree diagram above, the probability distribution table of R is r 7 P(R = r) ER r PR r 7 all r (b) From the question, the values X can take are 0,,,,, 0 and. Consider the table of outcome for X: To get : (, ), (, ), (, ), (,) To get : (, ), (, ), (, ), (,) To get : (, ), (, ) To get : (, ), (, ) To get 0: (, ), (, ) To get : (, ) Thus, we have the following probability distribution table: Hence, E x 0 0 P(X = x) X x P X x all x Page of

19 Topical Revision Package III Suggested Solutions H Mathematics (7) E X x P X x all x X X X Var E E 7 7. Note: R can only take the values 0, and since there are only red balls in the bottle. 7 7 PR PR PR r P(R = r) 7 PR PR PR (Shown) 7 7 ER r PR r 0 all r (Shown) 7 7 Var R E R 0 7 Note: M = R 7 7 E P all j Note: J j J j j P(J = j) 7 7 Page of

20 Topical Revision Package III Suggested Solutions H Mathematics (7) Thus, 7 7 Var J E J 7 EM J EM EJ ER EJ 0 Var Var Var Var Var M J M J R J E X xp X x all x E X x P X x all x 7 Thus, Var X E X E X Let Y = X X. 7 (Shown) y 0 P(Y = y) E P 0 all y Thus, Y y Y y (a) (b) True. Since W values is the same as the Y, their respective expectation will be the same. False. Since there are more W values than Y values, their respective variance will not be the same. Page 0 of

21 Topical Revision Package III Suggested Solutions H Mathematics (7). Let Y denote the score of the first die when thrown once. E P all y Y y Y y Y Y Var E Let W denote the score of the second die when thrown once. E P all w W w W w For variance, since Y EY W EW, then E Y EY E W EW. Thus, Var W Var Y. 7 Note: X = Y + Y + W, where Y and Y are independent observations of Y. (a) Var X Var Y Var Y Var W Var Y (Shown) 7 7 E E E E E E (b) Note: X Y Y W Y W 7 P P X Thus, X 7 7 P X P X 7 7 P X P X X X P.0 P. X X P P Page of

22 Topical Revision Package III Suggested Solutions H Mathematics (7) (Shown) Page of

Topic 5: Probability. 5.4 Combined Events and Conditional Probability Paper 1

Topic 5: Probability. 5.4 Combined Events and Conditional Probability Paper 1 Topic 5: Probability Standard Level 5.4 Combined Events and Conditional Probability Paper 1 1. In a group of 16 students, 12 take art and 8 take music. One student takes neither art nor music. The Venn