Subtopic : 10.1 Events and Probability

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Presentation transcript:

Subtopic : 10.1 Events and Probability LECTURE 1 OF 6 TOPIC 10 : PROBABILITY Subtopic : 10.1 Events and Probability

LEARNING OUTCOMES: At the end of the lesson, students should be able to (a) understand the concept of random experiments, outcomes, sample space, events and random selections (b) relate events as sets and use the basic set operations to define outcome of events

(c) understand exclusive and exhaustive events (d) define probability and understand some basic laws of probability

Suppose that there are a blue pen, a black pen and a grey pen. What’s the chance of choosing a black pen? Ans : Thus, the probability of choosing a black pen is

The Definition of Basic Terms in Probability (i) A random experiment ~ a process leading to at least two possible outcomes with uncertainty as to which they will occur. (ii) Basic outcomes ~ the possible outcomes of a random experiment (iii) Sample Space ~ the set of all basic outcomes ( denoted by S )

The result will be either a “ head ” or a “ tail ” Random Experiment Example 1 Tossing a fair coin The result will be either a “ head ” or a “ tail ” Basic Outcomes Sample Space S = { H ,T } H ~ head , T ~ tail

(iv) Event ~ is a subset of basic outcomes from the sample space. Example 2 Rolling a die the basic outcomes are the numbers 1, 2, 3, 4, 5 and 6. 2 4 6 Thus S = { 1, 2, 3, 4, 5 ,6 } 1 3 5 If A = { 1, 3, 5 } Then A is an event of getting odd numbers

Example 3 List a sample space when two dice are tossed or a die is tossed twice . S = { (1,1) , (1,2) , (1,3) , (1,4) , (1,5) , (1,6) (2,1) , (2,2) , (2,3) , (2,4) , (2,5) , (2,6) (3,1) , (3,2) , (3,3) , (3,4) , (3,5) , (3,6) (4,1) , (4,2) , (4,3) , (4,4) , (4,5) , (4,6) (5,1) , (5,2) , (5,3) , (5,4) , (5,5) , (5,6) (6,1) , (6,2) , (6,3) , (6,4) , (6,5) , (6,6) } Thus n(S) = 36

Let A be an event of getting the sum of two numbers is 6 Thus A = { (1,5) , ( 2,4) , (3,3) , (4,2) , (5,1) } n(A) = 5 Let B be an event of getting the sum of two numbers is a multiple of 5 Thus B = { (1,4) , ( 4,1) , (2,3) , (3,2) , (5,5) , (4,6) , (6,4) } n(B) = 7

Mutually Exclusive Events Two events A and B are called mutually exclusive if there is no intersection between the two events. A  B = A B

Let A : getting even numbers B : getting odd numbers A={2,4,6} Example 4 Rolling a die Let A : getting even numbers B : getting odd numbers A={2,4,6} B={1,3,5} Therefore, A  B = A B 4 6 3 5

Exhaustive Events Two events A and B are said to be exhaustive events if A1 A2 A3 A4 A5 A6 A7 A8

Example 5 (i) Let S={1,2,3,4,5,6,7,8,9,10}. If A={1,2,3,4,5,6} and B ={5,6,7,8,9,10}, then n (A  B) =n (S). Therefore, A & B are exhaustive events.

(ii) Let S={1,2,3,4,5,6}. If A={2,4,6} and B={1,3}, then n (A  B) ≠ n (S). Therefore, A and B are not exhaustive events.

The Probability Of An Event Definition The probability of an event A occurring is denoted by P(A) ,where P(A) = =

P(A) satisfies the following conditions If P(A) = 1 , => the event A is a sure event If P(A) = 0 , => the event A is an impossible event

The complement of an event A is denoted as with where P( ) or P(A’) means the probability that A doesn’t happen

Example 6 Two dice are tossed, find the probability the sum of the two numbers is 8 the sum of the two numbers is a prime number Solution : (a) Let A be the event the sum of the two numbers is 8 A = { (2,6) , (6,2) , (3,5), (5,3) , (4,4) } n(A) = 5 , n(S) = 36 Thus P(A) =

(b). Let B be the event of getting the (b) Let B be the event of getting the sum of the two numbers is a prime number. B = { (1,1) , (1,2) , (1,4) , (1,6) , (2,1), (2,3) , (2,5) , (3,2) , (3,4) , (4,1) ,(4,3) , (5,2) , (5,6), (6,1) , (6,5) } n(B) = 15 , n(S) = 36 Thus P(B) =

Example 7 Two fair coins are tossed simultaneously. Find the probability of getting (a) exactly two heads (b) at least one head

Solution : Die 2 Outcomes: Die 1 H H , H H T H , T H T , H T T T , T

(a) S={HH,HT,TH,TT} Let A be the event of getting 2 heads n(S) = 4 A = {HH} Thus P(A) = (b) Let B be the event of getting at least one head B = { HH,HT,TH} Thus P(B) =

Venn Diagram A Venn Diagram can also be used to solve probability problems Example 8 There are 100 form six students , of whom 20 are studying biology, 15 are studying chemistry and 8 are studying both Biology and Chemistry. We can illustrate this in a Venn diagram.

P( studying both Biology and Chemistry ) Venn diagram B ~ Biology C B C ~ Chemistry 12 8 7 73 P( studying both Biology and Chemistry ) P( B ∩ C ) = =

P(study either Biology or Chemistry or both ) = = 0.27 P(B U C)

Exercises 1. On a single toss of one die, find the probability of obtaining (a) a 4 (b) an odd number (c) an even number (d) a number less than 4 (e) a number greater than 4 (f) an odd or an even number

2. In a junior school class of 28 pupils, 2. In a junior school class of 28 pupils, 7 are in both a sports team and the school band. There are 16 pupils involved in sports teams and 10 in the school band. Find the probability that a pupil chosen at random. (a) is only in the school band (b) is in either a sports team or the school band (c) is in neither a sports team nor the school band

Conclusion 1. Mutually Exclusive Events A  B = Ø 2. Exhaustive Events n(A  B) = n(S)

3. Probability of Event A P(A) = 0 ≤ P(A) ≤ 1 4. Complement of Event A