ACJC 2025 Probability Lecture Notes

1 9 PROBABILITY SYLLABUS • Learn to apply addition and multiplication of probabilities • Understand mutually exclusive events and independent events • Use tables of outcomes, Venn diagrams, tree diagrams and permutation and combination technique to calculate probabilities. • Calculate conditional probabilities in simple cases • Use of: - P( ) 1 P( )AA =− - P( ) P( ) P( ) P( )A B A B A B = + −  - P( )P( | ) P( ) ABAB B =

ACJC 2025/26 H2 Mathematics (9758) 2 LECTURE 4 Lesson Outline • Guided self -practice questions of how and why each solving approach is applied 4 SELF-PRACTICE QUESTIONS WITH GUIDED SOLUTIONS After attempting the guided questions in this section, you may go to the Maths Google Site at https://sites.google.com/acjc.edu.sg/acjcmaths/jc1- class-of-2025/class-of-2025-h2-maths for complete solutions. Guided Self-Practice Question 1 Two boxes, one white and one black, are used in a game. The boxes contain balls labelled ‘2’, ‘5’ and ‘10’. The balls are identical except for their numbers. The table below shows the number of balls in each box. Number of balls labelled ‘2’ ‘5’ ‘10’ White box 5 2 1 Black box 5 1 2 In the first round of the game, a player draws 3 balls randomly from the white box. If the numbers on the 3 balls add up to 9 or more, the player enters the second round of the game to draw another 3 balls randomly from the black box. (i) Find the probability that the player draws 3 different numbers in the first round. [1] (ii) Show that the probability that sum of the numbers add up to 12 and 6, in the first and second round respectively, is 25 1568 . Hence, find the probability that the sum of the numbers drawn from the two rounds adds up to 18. [4] [RVHS/2019/II/Q6]

9 Probability 3 Solution (i) Method 1: Using “Combination” Number of ways to draw 3 different numbers in the first round =5  2  1 = 10 ways Total number of ways to select any 3 numbers 8 56 ways3 == P(player draws 3 different numbers in the first round) 10 56 5 28 = = Method 2: Using “Fractions” Possible cases are: ‘2’, ‘5’, ‘10’ ‘2’, ‘10’, ‘5’ ‘10’, ‘5’, ‘2’ P(for the case '2', '5', '10') 5 2 1 8 7 6=   5 1 2P(for the case '2', '10', '5') 8 7 6=   125P(for the case '10', '5', '2') 8 7 6=   Since each case gives the same answer, thus, P(player draws 3 different numbers in the first round) 5 2 1 5 3 !=8 7 6 28   Since all the 3 objects selected has to be distinct, using ( ) ( ) ( )521 1 1 1 , is also acceptable. However, it is not encouraged to use the method of always ‘choosing 1’, i.e. ( ) ( ) ( ) 1 1 1 a b k , since in most of the questions, the objects selected usually are not all distinct. In probability, we are calculating ‘number of possible ways’ out of ‘total number of ways without any restriction’. It is necessary to find the total number of cases. To begin, list out a few possible cases. Make use of ‘factorial’ to find the number of ways for n distinct “objects” to permutate. In this case, there are 3 distinct “objects”, ‘2’, ‘5’ and ‘10’. Hence, there are 3! ways, thus 6 cases. The cases are considered different since the orders are different.

ACJC 2025/26 H2 Mathematics (9758) 4 (ii) Method 1: Using “Combination” Drawing 3 balls from the white box, to obtain a total sum of 12, there should be one ‘2’ and two ‘5’s. Similarly, from the black box, to obtain a total sum of 6, there should be three ‘2’s. Number of ways to obtain a sum of 12 from the first round 52 12    =        Number of ways to obtain a sum of 6 from the second round 5 3 =   Required probability P(sum from first round is 12 sum from s econd round is 6)= and 5 2 5 1 2 3 88 33 25 (shown)1568                 =             = Hence, P(sum of the numbers drawn adds up to 18) P(sum of each of 1st and 2nd round is 9) P(sum of 1st round is 12 and sum of 2nd round is 6) =+ 5 2 5 1 25 1252 1 2 1 88 1568 1568 33                       =  + =             As not all the 3 objects selected are distinct, ( ) ( ) ( )521 1 1 1 will lead to multiple counting. For selection of one ‘2’ and two ‘5’s, use ( ) ( )52 12 . Since the 3 objects selected are NOT distinct, ( ) ( ) ( )5 4 3 1 1 1 will lead to multiple counting. To select three ‘2’s, use ( ) 5 3 . In probability, we are calculating ‘number of possible ways’ out of ‘total number of ways without any restriction’. Since both rounds need two ‘2’s and one ‘5’, the 3 ‘objects’ selected in each round are not all distinct, ( ) ( ) ( )5 4 2 1 1 1 will again lead to multiple counting. To select two ‘2’s and one ‘5’, use ( ) ( )52 21 . Since the 3 objects selected are not distinct, ( ) ( )54 11 will lead to multiple counting. To select two ‘2’s, use ( ) 5 2 .

9 Probability 5 Method 2: Using “Fractions” Possible cases are: ‘2’, ‘5’, ‘5’ ‘5’, ‘2 , ‘5’ ‘5’, ‘5’, ‘2’ 215P(for the case '2', '5', '5' from the white box)= 8 7 6 2 1 5P(for the case '5', '5', '2' from the white box) 876=   Since each case gives the same answer, thus, P(sum of numbers from 1st round is 12) 2 1 5 3 ! 8 7 6 2 ! =    Now, for obtaining three ‘2’s from the black box, 5 4 3P(for the case '2', '2', '2') 876=   Since all ‘2’s are the same, there are no other cases. Required probability P(sum from first round is 12 sum from s econd round is 6)= and 2 1 5 3 ! 5 4 3 25 8 7 6 2 ! 8 7 6 1568 =       = (shown) Hence, P(sum of the numbers drawn adds up to 18) P(sum of each of 1st and 2nd round is 9) P(sum of 1st round is 12 and sum of 2nd round is 6) =+ 5 4 2 3 ! 5 4 1 3 ! 25 8 7 6 2 ! 8 7 6 2 ! 1568 125 1568 =        + = ■ Listing seems easy here since there are only 3 cases. Alternatively, using factorial, with 2 repeated ‘5’s out of the 3 numbers, there are 3! 2! ways. Again, listing seems easy here since there are only 3 cases. Alternatively, using factorial, with 2 repeated ‘2’s out of the 3 numbers, there are 3!/2! ways for both 1st and 2nd rounds.

ACJC 2025/26 H2 Mathematics (9758) 6 Guided Self-Practice Question 2 A computer game consists of at most 3 rounds. The game will stop when a player clears 2 rounds or does not clear 2 consecutive rounds. The probability that a player clears round 1 is 0.6. The conditional probability that the player clears round 2 given that he clears round 1 is half the probability that he clears round 1. The conditional probability that the player clears round 2 given th at he does not clear round 1 is the same as the probability that he clears round 1. (i) Find the probability that a player plays 3 rounds. [1] (ii) Find the probability that a player clears round 1, given that he does not clear round 2. [2] (iii) The total probability that a player plays 3 rounds and clears round 3 is 0.2. Find the probability that a player clears exactly 2 rounds. [2] [TMJC/2019/II/Q6(i)-(iii)] Solution (i) Fill in the boxes with the relevant probabilities in the probability tree in the next page. (Recall: For each pair of branches, the probabilities add up to 1…) ( ) ( ) ( ) P a player plays 3 rounds P clears round 1 but doesn't clear round 2 P doesn't clear round 1 but clears round 2 = + = (0.6)(0.7) + (0.4)(0.6) = 0.66 0.6 0.4 0.4 0.6 0.7 0.3 Round 1 Round 2 Round 3* Clear Doesn’t Clear Clear Doesn’t Clear Clear Doesn’t Clear Clear Doesn’t Clear Clear Doesn’t Clear