SAJC 2021 JC1 H1 Maths Final Exam concepts

Page 1 2021 JC1 H1 Maths FE level up concepts Holiday REVISION  Please present all your answers on this copy as it will be submitted to your tutor when Term 1 starts in 2022.  If in doubt, please read your lecture notes, if still further in doubt, put a note under ‘Learning Point’ so that you remember to clarify your doubts with your tutor. Where possible, answers should be left as fraction. Question Solutions Learning Point 1. Permutation and Combination (Arrangement) (i) Three friends, Andy, Ben and Charlie, attend a graduation ceremony with their parents. How many ways can the nine people be seated in a row if each family must be seated together? Ans: 1296 Arrange within family = 3! Arrange 3 families = 3! 3! 3! Total arrangement = 3! 3! 3! 3! 1296    (ii) How many ways can the nine people be seated in a row if each set of parents must be seated together? Ans: 5760 Arrange within each set of parents = 2! Arrange 3 sets of parents and 3 friends = 6! Total arrangement =   3 2! 6! 5760 (iii) How many ways can the nine people be seated in a row if the 3 friends must be seated together? Ans: 30240 Arrange within the 3 friends (1 unit) = 3! Arrange 1 unit and 6 parents = 7! Total arrangement = 7! 3! 30240

Page 2 (iv) How many ways can the nine people be seated in a row if the 3 friends wants to be separated? Ans:151200 Arrange all 6 parents first (no restriction) = 6! SLOT the 3 friends into the 6 parents = 7 3P Total arrangement = 7 3 6! 151200P  (v) How many ways can the nine people be seated in a row if each set of parents must be seated together AND the 3 friends wants to be separated? Ans:1152 Arrange within each set of parents =   3 2! Arrange 3 sets of parents = 3! SLOT the 3 friends into the 3 sets of parents = 4 3P (Think carefully here!) Total arrangement =   3 4 32! 3! 1152P   (vi) How many ways can the nine people be seated in a row if Andy must take the first seat, Ben must take the last seat and each set of parents must be seated together? Ans:192 Arrange within each set of parents =   3 2! Arrange Charlie and 3 sets of parents = 4! Total arrangement =   3 2! 4! 192

Page 3 2. Permutation and Combination (Combination) In a college student committee, there are 4 Administrators, 5 Liaison officers and 6 Group Leaders. They are to form an Ex-Co that consists of a President, a Vice-President and a Secretary. (i) How many selections are there to form the Ex-Co? Ans 2730 Total selection (no restriction) = 15 3P 2730 (ii) How many selections are there to form the Ex-Co if a President must be from the Administrators, a Vice- President must be from the Liaison Officers and a Secretary must be from the Group Leaders? Ans : 120 To select a President from Administrators = 4 1C To select a Vice President from the Liaison Officers = 5 1C To select a Secretary from the Group Leaders = 6 1C Total selections = 4 5 6 1 1 1C C C 120   (iii) How many selections are there to form the Ex-Co if at least 1 of them must be from the Administrators? Ans: 1740 Method 1 (Complementary) No restriction = 15 3P 2730 Nobody from the Administrators = 11 3P 990 Total selections = 2730 990 1740 Method 1 (Consider cases though not recommended) Case 1: 1 from the Administrators, the other 2 from others = 4 11 12C C 3! 1320   Case 2: 2 from Administrators, the other from others = 4 11 21C C 3! 396   Case 3: 3 from Administrators = 4 3P 24 Total selections =1320 + 396 + 24 = 1740

Page 4 (iv) How many selections are there to form the Ex-Co if at least 2 of them must be from the Administrators and none must be from Group Leaders? Ans: 204 Consider cases (only) Case 1: 2 from Administrators, the other 1 from Liaison Officers = 45 21C C 3! 180   Case 2: all 3 from the Administrators = 4 3P 24 Total selections = 180 + 24 = 204 (v) How many selections are there to form the Ex-Co if an Administrator is already made the President and only another one must be from the Administrators? Ans: 66 one Ex-Co must be from the Administrators = 3 1C 1 Ex-Co from others = 11 1C Total selections = 3 1C 11 1C 2! 66

Page 5 3. Probability In an egg production factory, eggs are packed into cases on 4 different production lines, 1 2 3L ,L ,L and 4L . The proportion of packed eggs that come from production lines 1 2 3L ,L ,L and 4L are 7 ,20 1 ,5 3 20 and 3 10 respectively. Records show that a small percentage of eggs are not packed properly for sale. 1% from 1L , 3% from 2L , 3% from 3L and p% from 4L . A case is chosen at random. Draw a tree diagram to represent this information. (i) Given that the probability of the chosen case is faulty is 0.02. Find p. Ans: p = 2  P a case is faulty 0.02 7 1 1 3 3 3 3 0.0220 100 5 100 20 100 10 100 30.014 0.021000 3 0.0061000 36 2 p p p p p                               L1 L2 L3 L4 1 5 3 20 3 10 7 20 P P’ P 3 100 3 100 1 100 P P’ P P’ p 100 99 100 97 100 97 100 1 100 p P: packed properly P’: not packed properly P’

Page 6 ii) Find the probability that the case is not faulty and is from production line 1L . Ans : 693 2000  1 7 99P a case is from L and it is not faulty 20 100 6930.3465 or 2000    (iii) Find the probability that the case is not faulty or is from production line 1L . Ans : 0.9835  1P a case is from L or it is not faulty 7 98 693 20 100 2000 0.9835     (iv) Find the probability that a case that is not faulty is from 1L . Ans : 99 280       1 1 P a box is from L |a box is not faulty P a box is from L and is not faulty P a box is not faulty 0.3465 1 0.02 99 280     (v) Ten randomly cases are chosen. Find the probability that none are faulty. Ans :0.817 Required probability =   10 0.98 = 0.817

Page 7 4. Binomial Distribution From a production line making toys, a fixed number of toys are inspected for faults and that 10% of the toys are faulty. (i) State, in context, two assumptions needed for the number of faulty toys found to be well modelled by a binomial distribution. (Looking for keywords): (event+independence; probability+constant) Assunption 1: event that a faulty toy selected is independent of another faulty toy selected Assumption 2: Probability that a toy is faulty is a constant at 0.1 for each toy. (ii) 8 toys were selected. Find the probability that less than 4 toys are faulty. Ans: 0.995 Let X be no of faulty toys out of 8._ X~B(___8____ , ___0.1_______) Less than 4 means: P( 4) P( 3) 0.99498 0.995XX     (iii) 8 toys were selected. Find the probability that at least 3 toys are faulty. Ans: 0.0381 at least 3 toys are faulty means: P( 3) 1 P( 2) 0.0381XX     (iv) 8 toys were selected. Find the probability that between 3 to 5 (inclusive) toys are faulty. Ans: 0.0381 P(3 5) P( 5) P( 2) 0.03807 0.0381X X X       

Page 8 (v) A box is packed with 8 toys. There are 10 such boxes. Find the probability that there are at most 8 such boxes with less than 4 toys are faulty. Ans: 0.001104 10 boxes with 8 toys in each box Let W be no of boxes with less than 4 toys that are faulty out of 10. W~B(___10____ ,