2025_Y6T2W1Mathfocus_Probability_Soln

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 6 _____________________________________ Y6 H2 Math Term 2 Week 1 Revision: Probability Page 1 of 4 Term 2 Week 1 Revision Topic: Permutations and Combinations and Probability 1 JPJC Prelim 9758/2020/02/Q6 The eleven letters in the word CORONAVIRUS are rearranged to form ‘words’ which may not make sense. Find the probability that (i) the two ‘R’s are together; [3] (ii) the vowels (O, A, I, U) are separated; [3] (iii) either the first letter is a ‘C’ or the last letter is an ‘S’ or both. [3] Solution (i) No. of ways where the two ‘R’s are together = 10! 18144002! Required probability = 1814400 1814400 2 11! 9979200 11 2!2!   or 0.182 (ii) No. of ways to arrange the consonants = 6! 3602! No. of ways to slot in the vowels = 7 5 5! 12602!C  Required probability = 360 1260 453600 1 11! 9979200 22 2!2!    (or 0.0455) (iii) No of ways that the first letter is a C or last letter is an S10! 2 18144002!2!   No of ways that the first letter is a C and last letter is an S9! 907202!2!  Required probability 1814400 90720 1723680 19 11! 9979200 110 2!2!    (or 0.173)

Raffles Institution H2 Mathematics 2025 Year 6 _________________________________________________________________________________________ _____________________________________ Y6 H2 Math Term 2 Week 1 Revision: Probability Page 2 of 4 2 CJC Prelim 9758/2020/02/Q6 A bag contains 4 red counters and 6 blue counters. 4 counters are drawn from the bag at random, without replacement. (a) Calculate the probability that (i) all the counters drawn are blue, [1] (ii) at least 3 blue counters are drawn, [2] (iii) at least 1 counter of each colour is drawn, [2] (iv) at least 3 blue counters are drawn, given that at least 1 of each colour is drawn. [2] (b) State with a reason whether or not the events “at least 3 blue counters are drawn” and “at least 1 counter of each colour is drawn” are independent. [1] Solution (a)(i) P (all counters blue) 6 4 10 4 1 14 C C  or 6 5 4 3 10 9 8 7   (a)(ii) Case 1 : P(3 blue,1 red) 6 4 3 1 10 4 8 21 C C C   or 6 5 4 4 4! 10 9 8 7 3!    Case 2 : P (all blue) 1 14 P(at least 3 blue) = P(3 blue, 1 red) + P(all blue) 8 1 19 21 14 42   (a)(iii) P (all counters red) 4 4 10 4 1 210 C C  or 4 3 2 1 10 9 8 7   P(at least 1 counter of each colour is drawn) = 1 – P(all blue) – P(all red) 1 1 971 14 210 105    (a)(iv) P(at least 3 blue | at least 1 of each colour) P(at least 3 blue at least 1 of each colour) P(at least 1 of each colour)  P(3 blue,1 red) P(at least 1 of each colour)  8 4021 97 97 105   (b) Since P(at least 3 blue | at least 1 of each colour) = 40 19 97 42  P (at least 3 blue), the two events are not independent.

Raffles Institution H2 Mathematics