ACJC 2025 Probability Summary

Anglo-Chinese Junior College 2025 H2 Mathematics 9758: Probability / Summary / Page 1 of 4 SUMMARY: Probability For uniform sample space number of outcomes in P( ) total number of outcomes AA  Combination of events A and B This means outcomes satisfying both A and B, i.e. AÇ B  To calculate P( )A BÇ , determine the actual outcomes in AÇ B A or B This means all outcomes in A or B or both, i.e. A B  To calculate P( )A B , usually easier to use P( ) P( ) P( ) P( )A B A B A B    Ç  Formula can be extended to P( )A B C  , P( ) P( ) P( ) P( ) P( ) P( ) P( ) P( ) A B C A B C A B B C C A A B C       Ç  Ç  Ç  Ç Ç Conditional Probability P( | )A B is the probability of A given that B occurred P( )P( | ) P( ) A BA B B Ç  Can be found by ‘reduced sample space’ method Relationship of Two Events Mutually Exclusive No intersection of events A and B, i.e. P( ) 0A BÇ  .  Can be seen from Venn Diagram A B A B A B

Anglo-Chinese Junior College 2025 H2 Mathematics 9758: Probability / Summary / Page 2 of 4 Independence Two events A and B are independent if the probability of A is not affected by whether B occurs or not.  It cannot be seen on Venn diagram  A and B are independent if and only if P( ) P( ) P( )A B A BÇ  or P( | ) P( )A B A  A and B independent implies A and B are independent, etc. Methods and Approaches Students should remember the preferred method or approach for different types of questions. (i) Venn Diagram (ii) Tree Diagram (iii) P & C / Probability Method Example (Venn Diagram) For events A and B, it is given that 11P( ) 20A  and 1P( ) 2B  . (i) Find the greatest and least possible values of P( )A BÇ . It is given in addition that 7P( | ) 9B A   . (ii) Find P( )A B . (iii) Determine if A and B are independent events. Justify your answer. Solution (i) P( ) 1A B  P( ) P( ) P( ) 1 1P( ) P( ) P( ) 1 20 A B A B A B A B   Ç  Ç     min 1P( ) 20A BÇ  The biggest P( )A BÇ occurs when B is a subset of A. Hence max 1P( ) 2A BÇ  .

Anglo-Chinese Junior College 2025 H2 Mathematics 9758: Probability / Summary / Page 3 of 4 (ii) P( ) 7P( | ) P( ) 9 B AB A A Ç    7 7 9 7P( ) P( ) 9 9 20 20B A A     Ç         7 11 9P( ) P( ) P( ) 20 20 10A B B A A   Ç     (from Venn diagram) (iii) 7 1P( | ) P( ) 9 2B A B    , therefore B and A are not independent and hence B and A are not independent. OR Since 1 9 9 7P( ) P( ) P( ) 2 20 40 20B A B A        Ç     , therefore B and A are not independent and hence B and A are not independent. OR 1 7 3P( ) P( ) P( ) 2 20 20A B B B A Ç   Ç    Since 11 1 11 3P( ) P( ) P( ) 20 2 40 20A B A B       Ç     , A and B are not independent.

Anglo-Chinese Junior College 2025 H2 Mathematics 9758: Probability / Summary / Page 4 of 4 Example (Tree Diagram) John and Peter play a game of chess. It is equally likely for either player to make the first move. If John makes the first move, the probability of him winning the game is 0.3 while the probability of Peter winning the game is 0.2. If Peter makes the firs t move, the probability of him winning the game is 0.5 while the probability of John winning the game is 0.4. If there is no winner, then the game ends in a draw. (i) Find the probability of a game ending in a draw. (ii) Find the probability that Peter made the first move given that he won the game. Solution (i)  P(draw) 0.5)(0.5 (0.5)(0.1) 0.3   (ii) P(Peter first move | Peter won)          P(Peter first move and Peter won) P(Peter won) 0.5 0.5 5 70.5 0.2 0.5 0.5     Example (P&C/Probability Method) A bag contains 5 red, 3 green and 4 blue balls. Three balls are drawn randomly from the bag (without replacement). Find the probability that (i) all three balls are of different colour, (ii) all three balls are of the same colour, (iii) there are 2 green balls and 1 blue ball, (iv) there are more green balls given that there is at least 1 blue ball drawn. Solution (i)   5 3 4 1 1 1 12 3 3P all different 11 C C C C  or 5 3 4 3 3!12 11 10 11    (ii)  P all same      P RRR P GGG P BBB   5 3 4 3 3 3 12 3 3 44 C C C C    or 5 4 3 3 2 1 4 3 2 12 11 10 12 11 10 12 11 10        3 44 (iii)   3 4 2 1 12 3 3P 2 green 1 blue 55 C C C  or 3 2 4 3! 3 12 11 10 2! 55    (iv)  P more green | 1 blue         8 3 12 3 3 55P more green and 1 blue P 2 green and 1 blu e 3 P 1 blue P 1 blue 41 1 C C        John Peter John wins Peter wins John wins Peter wins 0.5 0.5 0.3 0.5 0.2 0.4 Draw 0.5 Draw 0.1