SAJC 2021 Ch2 Probability Teacher version

SAJC 2021 JC1 H1 Mathematics Chapter 2: Probability 1 Chapter 2 (Statistics): Probability (Teacher’s copy) Objectives At the end of the chapter, you should be able to (a) understand that the probability of an event measures how likely the event will occur (b) construct a table of outcomes to calculate probabilities, and understand that the total probability of all possible outcomes is equal to 1 (c) calculate probabilities using addition and multiplication principles (d) use a Venn diagram to interpret probabilities such as P( )A , P( )AB , P( )AB , P( | )AB (e) understand the meaning of mutually exclusive events, and recognize events that are, or are not, mutually exclusive through practical examples; and use the result P( ) P( ) P( )A B A B   where A and B are mutually exclusive (f) understand the meaning of independent events, and use the result P( ) P( )P( )A B A B , where A and B are independent. (g) construct a tree diagram and use it to interpret and calculate probabilities, including probabilities of combined events and conditional probabilities. Contents 2.1 Basic Rules and Definitions 2.1.1 Basic Definitions 2.1.2 Classical (Theoretical) Definition of Probability 2.1.3 Basic Probability Rules 2.2 Tools for Finding Probabilities 2.2.1 Venn Diagram 2.2.2 Tree Diagram 2.2.3 Table of Outcomes 2.3 Conditional Probability 2.4 Mutually Exclusive Events 2.5 Independent Events 2.6 A Mix of Probabilities "Life is a school of probability.” --Walter Bagehot

SAJC 2021 JC1 H1 Mathematics Chapter 2: Probability 2 2.1 Basic Rules and Definitions 2.1.1 Basic Definitions Definition Experiment 1 Experiment 2 An experiment or trial is a process that generates data. Throw a six-sided fair die. Toss a fair coin 2 times. An outcome is the result of a single trial of an experiment. Obtain a ‘2’ Obtain a ‘Head’, then a ‘Tail’ (HT) The sample space, S, of an experiment is the set of all possible outcomes. The number of outcomes in the sample space is denoted by n(S). {1, 2, 3, 4, 5, 6} n(S)= 6 {HH, HT, TH, TT} n(S)= 4 An event is a subset of the sample space. The number of outcomes in an event A is denoted by n(A). Let A be the event “a prime number is obtained”. A = {2, 3, 5} n(A)= 3 Let B be the event “2 heads is obtained”. B = {HH} n(B)= 1 Probability is the measure of how likely an event is to occur. AA S n( ) 3 1P( ) = = =n( ) 6 2 1 4 BB S n( )P( ) = =n( ) 2.1.2 Classical (Theoretical) Definition of Probability For equally likely outcomes from a finite sample space S, the probability of an event A is defined as P(A) Number of Ways Event can occur n( ) Total Number of Possible Outcomes n( ) AA S Example 1 A playing card is to be drawn at random from a pack of 52 cards. Find the probability that (i) it will be red, (ii) it will be a heart or an ace. Solution: (i) Event A : a red card is obtained n(A) = 26 n(S) = 52 (ii) Event B: card is a heart or ace n(B) = total number of cards that will be a heart or an ace = 13 + 3 = 16 n(S) = 52

SAJC 2021 JC1 H1 Mathematics Chapter 2: Probability 3 P( AB ) P(card will be red) = P(A )= 26 1 52 2 P(card will be a heart or an ace) = P(B) = 16 4 52 13 2.1.3 Basic Probability Rules Suppose there are 2 events, A and B, 1. 0  P(A)  1 If event A is impossible to happen, then P (A) = 0. If event A is an absolute certainty to happen, then P (A) = 1. A lot of the time, you’ll be dealing with probabilities somewhere in between. 2. Complement of A means A does not happen. P( A ) = 1 P( )A 3. Union ( A B ) and Intersection ( AB ) P( A B ) means the probability of A or B (or both) happening. P( AB ) means the probability of both A and B happening. P( A or B) = P( AB ) = P( ) P( ) P( )A B A B   A A’ P( A B ) B A P( AB )

SAJC 2021 JC1 H1 Mathematics Chapter 2: Probability 4 2.2 Tools for finding Probabilities 2.2.1 Venn Diagram A diagram that shows all possible logical relations between a finite collections of sets. Shade the Venn Diagram to produce these results: 1. P(( A B )’) = P( ''AB ) 2. P( ( ) 'AB ) = P( ''AB ) Note that case 1 & 2 above are representations of De Morgan’s Laws. 3. P( 'AB ) = P( A ) − P( AB ) 4. P ( 'AB ) Example 2 Events A and B are such that P( ) 0.3, P( ) 0.4 and P( ) 0.1A B A B    . Find (i) P( )AB (ii) P( )AB  (iii) P( )AB (iv)  P ( )AB  Solution: (i) P( ) P( ) P( ) P( ) 0.3 0.4 0.1 0.6A B A B A B         (ii) P( ) P( ) P( ) 0.3 0.1 0.2A B A A B       (iii) P( ) 1 P( ) 1 0.6 0.4A B A B       (iv)    P ( ) 1 P 1 0.1 0.9A B A B       A B A B A B B B A A

SAJC 2021 JC1 H1 Mathematics Chapter 2: Probability 5 Example 3 Analysis of the results of a certain group of students who had taken examinations in both Mathematics and Economics produced the following information: 75% of the students passed Mathematics, 70% passed in Economics and 60% passed both subjects. Find (i) the percentage of students who passed at least 1 subject; (ii) the percentage of students who had passed exactly one of two subjects; (iii) the probability that a student failed both subjects. Solution: Let Event A: a student passed Mathematics; Event B: a student passed Economics. Given: P(A) = 0.75 , P(B) = 0.7 , P( )AB = 0.6 (i) P(passed at least one subject) = P( )AB = 0.75+0.7  0.6 = 0.85 Therefore. 85 % passed at least 1 subject. (ii) P(passed exactly one of the two subjects) = P( ') P( ' ) 0.15 0.1 0.25 A B A B      Therefore, 25% of the students had passed exactly one of the two subjects. (iii) P(a student failed both subjects) = P( ) 1 P( )A B A B    = 1  0.85 = 0.15 Example 4 In a race in which there are no ties, the probability that John wins is 0.3, the probability that Paul wins is 0.2 and the probability that Mark wins is 0.4. Find the probability that (a) John or Mark wins, (b) John or Paul or Mark wins, (c) someone else wins. Solution: (a) P(John or Mark wins) = P(John wins)+ P(Mark wins) = 0.3 + 0.4 = 0.7 (b) P(John or Paul or Mark wins) = 0.3 + 0.2 + 0.4= 0.9 (c) P(someone else wins) = 1−0.9 =0.1 A B

SAJC 2021 JC1 H1 Mathematics Chapter 2: Probability 6 Exercise 1 1. Events C and D are such that   19P 30 C  ,   3P 5D  and   4P 5CD . Find  P CD . [Ans: 7 30 ] Solution:   2P 5D  P( ) P( ) P( ) P( ) 19 2 4 7P( ) P( ) P( ) P( ) 30 5 5 30 C D C D C D C D C D C D                2. For the sample space S, it is given that P( ) 0.5A  , P( ) 0.6AB and P( ) 0.2AB . Find (i) P( )B (ii) P( ' )AB (iii) P( ')AB (iv) P( ' ')AB [Ans: (i) 0.3 (ii) 0.1 (iii) 0.3 (iv) 0.4] Solution: (i) P( ) P( ) P( ) P( ) 0.6 0.5 P( ) 0.2 P( ) 0.3 A B A B A B B B          (ii) P( ' ) P( ) P( ) 0.3 0.2A B B A B      = 0.1 (iii) P P P 0 5 0 2 0 3( A B') ( A ) ( A B ) . . .       (iv) P( ' ') 1 P( ) 1 0.6 0.4A B A B       3. In a survey, 15 % of the participants said that they had never bought lottery tickets or premium bonds, 73% had bought lottery tickets and 49% had bought premium bonds. Find the probability that a participant chosen at random (a) had bought lottery tickets or premium bonds (or both), (b) had bought lottery tickets and premium bonds, (c) had bought lottery tickets only. [Ans (