Chapter 16 Probability 10312020 Probability by Chtan FYHS

Chapter 16 Probability 概率 10/31/2020 Probability by Chtan -- FYHS Kulai 1

Definition of Probability : If the possibility space S consists of a finite number of equally likely outcomes, then the probability of an event E written P(E) is defined as : 10/31/2020 Probability by Chtan -- FYHS Kulai 2

样本空间 S is called the sample space. n(S) is the number of the sample space. 事件 E is called the event. n(E) 10/31/2020 is the number of the event. Probability by Chtan -- FYHS Kulai 3

n(S)=a a-r S n(A)=r A 10/31/2020 Since, Probability by Chtan -- FYHS Kulai 4

Therefore, Note : 1. The probability of an event A is a number between 0 and 1 inclusive. 2. If P(A)=0, the event never occur. 3. If P(A)=1, the event is certain to occur. 10/31/2020 Probability by Chtan -- FYHS Kulai 5

e. g. 1 If a card is drawn from the clubs suit of a pack of cards, then P(card is red) = 0 P(card is black) = 1 Diamonds Spades Hearts Clubs 10/31/2020 Probability by Chtan -- FYHS Kulai 6

Let A’ denote the event “A does not occur”. Now, 10/31/2020 Probability by Chtan -- FYHS Kulai 7

or, 10/31/2020 Probability by Chtan -- FYHS Kulai 8

e. g. 2 A card is drawn at random from an ordinary pack of 52 playing cards. Find the probability that the card (a) is a seven, (b) is not a seven. 10/31/2020 Probability by Chtan -- FYHS Kulai 9

Soln : The possibility space S={the pack of 52 cards} and n(S)=52. Let A be the event “the card is a seven”, then n(A)=4. (a) P(A) = n(A)/n(S) = 4/52=1/13 (b) P(A’) = 1 - P(A)=1 10/31/2020 Probability by Chtan -- FYHS Kulai 1/13 = 12/1 10

e. g. 3 Compare the probability of scoring a 4 with one die and a total of 8 with two dice. 10/31/2020 Probability by Chtan -- FYHS Kulai 11

Soln : With one die, S={1, 2, 3, 4, 5, 6}; n(S)=6 Let A be the event “a 4 occurs”, then n(A)=1 Hence, 10/31/2020 Probability by Chtan -- FYHS Kulai 12

With two dice By permutation, the possible outcomes of two dice is 6 x 6=36 ways. Hence, n(S)=36 Let B be the event ‘the sum on the two dice is 8’. B={(2, 6), (6, 2), (3, 5), (5, 3), (4, 4)}, n(B)=5 P(B)=n(B)/n(S)=5/36 10/31/2020 Probability by Chtan -- FYHS Kulai 13

e. g. 4 Two fair coins are tossed. Find the probability that two heads are obtained. Soln : S={HH, HT, TH, TT}; n(S)=4 Let A be the event “two heads are obtained”. n(A)=1, 10/31/2020 P(A)=n(A)/n(S)=1/4 Probability by Chtan -- FYHS Kulai 14

Throw 2 dice, the possible outcomes. Illustrated by tree diagram. 1 2 3 4 5 6 10/31/2020 Probability by Chtan -- FYHS Kulai 15

If A and B are any two events of the same experiment such that and then Writing the result in set notation, 10/31/2020 Probability by Chtan -- FYHS Kulai 16

The Venn diagram : S B A r-t s-t t 10/31/2020 Probability by Chtan -- FYHS Kulai 17

10/31/2020 Probability by Chtan -- FYHS Kulai 18

e. g. 5 in a group of 20 adults, 4 out of the 7 women and 2 out of the 13 men wear glasses. What is the probability that a person chosen at random from the group is a women or someone who wears glasses? 10/31/2020 Probability by Chtan -- FYHS Kulai 19

Soln : Let W be the event “the person chosen is a woman” and G be the event “the person chosen wears glasses”. Now, 10/31/2020 Probability by Chtan -- FYHS Kulai 20

Mutually exclusive events 10/31/2020 Probability by Chtan -- FYHS Kulai 21

If an event A can occur or an event B can occur but not both A and B can occur, then the two events A and B are said to be mutually exclusive. 10/31/2020 Probability by Chtan -- FYHS Kulai 22

In this case , When A and B are mutually exclusive events, and 10/31/2020 Probability by Chtan -- FYHS Kulai 23

This is known as the addition law for mutually exclusive events. 10/31/2020 Probability by Chtan -- FYHS Kulai 24

Examples of mutually exclusive events: 1. A number is chosen from the set of integers from 1 to 10 inclusive. If A is the event “the number is odd” and B is the event “the number is a multiple of 4” then A and B are mutually exclusive events, as an event cannot be both odd and a multiple of 4. 10/31/2020 Probability by Chtan -- FYHS Kulai 25

2. Two men are standing for election as chairman of a committee. Let A be the event “Mr Smith is elected” and Y be the event “Mr Jones is elected”. Then A and Y are mutually exclusive events as both cannot be elected as chairman. 10/31/2020 Probability by Chtan -- FYHS Kulai 26

e. g. 6 In a race the probability that John wins is 1/3, the probability that Paul wins is ¼ and the probability that Mark wins is 1/5. Find the probability that (a) John and Mark wins, (b) neither John nor Paul wins. Assume that there are no dead heats. 10/31/2020 Probability by Chtan -- FYHS Kulai 27

Soln : We assume that only one person can win, so the events are mutually exclusive. (a) P(John or Mark wins)=P(John wins)+P(Mark wins) =1/3 + 1/5 = 8/15 (b) P(neither John nor Paul wins) = 1 – P(John or Paul wins) = 1 - (1/3 + ¼) = 1 – 7/12 = 5/12 10/31/2020 Probability by Chtan -- FYHS Kulai 28

e. g. 7 A card is drawn at random from an ordinary pack of 52 playing cards. Find the probability that the card is (a) a club or a diamond, (b) a club or a king. 10/31/2020 Probability by Chtan -- FYHS Kulai 29

Soln : (a) n(S)=52 Let C be the event “a club is drawn”, D be the event “a diamond is drawn”, K be the event “a king is drawn”. P(C)= n(C)/n(S) = 13/52 = 1/4 P(D)= n(D)/n(S) = 13/52 = 1/4 P(C U D)=1/4 + 1/4 = 1/2 10/31/2020 Probability by Chtan -- FYHS Kulai 30

(b) P(C)=13/52, P(K)=4/52 P(K n C)=P(king of club)=1/52 P(C U K)=P(C) + P(K) – P(C n K) =13/52 + 4/52 – 1/52 =16/52 =4/13 10/31/2020 Probability by Chtan -- FYHS Kulai 31

e. g. 8 Two ordinary dice are thrown. Find the probability that (a) at least one 6 is thrown, (b) at least one 3 is thrown, (c) at least one 6 or at least one 3 is thrown. Ans : (a) 11/36 10/31/2020 (b) 11/36 Probability by Chtan -- FYHS Kulai (c) 5/9 32

Soln : Let A be the event “at least one Six”, B be the event “at least one Three”. A={(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)} B={(1, 3), (2, 3), (3, 3), (4, 3), (5, 3), (6, 3), (3, 1), (3, 2), (3, 4), (3, 5), (3, 6)} n(S)=36 , n(A)=11, n(B)=11 (a) P(A)=n(A)/n(S)=11/36 (b) P(B)=n(B)/n(S)=11/36 10/31/2020 Probability by Chtan -- FYHS Kulai 33

(c) P(A U B)=P(A)+P(B)-P(A n B) = 2/36 Hence, P(A U B)= 11/36 + 11/36 – 2/36 = 20/36 = 5/9 10/31/2020 Probability by Chtan -- FYHS Kulai 34

Exhaustive events 10/31/2020 Probability by Chtan -- FYHS Kulai 35

If two events A and B are such that AUB=S then P(AUB)=1 and the events A and B are said to be exhaustive. 10/31/2020 Probability by Chtan -- FYHS Kulai 36

For example : (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 A U B = S A and B are exhaustive events. 10/31/2020 Probability by Chtan -- FYHS Kulai 37

(ii) Let S be the possibility space when an ordinary die is thrown. If A is the event “the number < 5” and B is the event “the number > 3” then the events A and B are exhaustive as A U B = S. 10/31/2020 Probability by Chtan -- FYHS Kulai 38

e. g. 9 Events A and B are such that they are both mutually exclusive and exhaustive. Find a relationship between A and B. Give an example of such events. 10/31/2020 Probability by Chtan -- FYHS Kulai 39

Soln : A and B are mutually exclusive then P(AUB)=P(A)+P(B) A and B are exhaustive then P(AUB)=1 Therefore, P(A)+P(B)=1 – P(A) 10/31/2020 Probability by Chtan -- FYHS Kulai 40

But P(A’)=1 - P(A) Hence P(B)=P(A’) i. e. B=A’ Similarly A=B’ Toss a coin. 10/31/2020 Probability by Chtan -- FYHS Kulai 41

"It's the little things that make the big things possible. Only close attention to the fine details of any operation makes the operation first class. " -- J. Willard Marriot 10/31/2020 Probability by Chtan -- FYHS Kulai 42

Conditional Probability 10/31/2020 Probability by Chtan -- FYHS Kulai 43

If A and B are two events and P(A) and P(B) are not equal to 0, then the probability of A , given that B has already occurred is written P(A|B) 10/31/2020 Probability by Chtan -- FYHS Kulai 44

Similarly, 10/31/2020 Probability by Chtan -- FYHS Kulai 45

Note : 10/31/2020 Probability by Chtan -- FYHS Kulai 46

Illustrating this by means of the Venn diagram, S n A B BA s-t An. B, t 10/31/2020 Probability by Chtan -- FYHS Kulai 47

This result is often written as : 10/31/2020 Probability by Chtan -- FYHS Kulai 48

Note : If A and B are mutually exclusive events then, as and , it follows that 10/31/2020 Probability by Chtan -- FYHS Kulai 49

e. g. 10 Given that a heart is picked at random from a pack of 52 playing cards, find the probability that it is a picture card. 10/31/2020 Probability by Chtan -- FYHS Kulai 50

Soln : We require 10/31/2020 Probability by Chtan -- FYHS Kulai 51

e. g. 11 When a die is thrown, an odd number occurs. What is the probability that the number is prime? 10/31/2020 Probability by Chtan -- FYHS Kulai 52

Soln : P(prime | odd)=P(prime n odd)/P(odd) The odd prime numbers are 3 and 5. 10/31/2020 Probability by Chtan -- FYHS Kulai 53

As We have It follows that Now Hence 10/31/2020 Probability by Chtan -- FYHS Kulai 54

e. g. 12 The probability of a bulb lifespan lasted for 5, 000 hours is ¾, lasted for 10, 000 hours is ½. There is a bulb still can be used after 5, 000 hours. What is the probability that it can be used till 10, 000 hours? 10/31/2020 Probability by Chtan -- FYHS Kulai 55

Soln : Let A be the event ‘the bulb can be used till 5000 hrs’, B be the event ‘the bulb can be used till 10, 000 hours’. Given P(A)=3/4 , P(B)=1/2 A bulb lasted till 10, 000 hours definitely also lasted 5, 000 hrs. Hence, An. B=B P(An. B)=P(B) 10/31/2020 Probability by Chtan -- FYHS Kulai 56

So, the question requires us to find the probability that a bulb ‘can be used till 10, 000 hrs given it has been used till 5, 000 hrs’. To find P(B|A)=? Therefore, 10/31/2020 Probability by Chtan -- FYHS Kulai 57

e. g. 13 A bag contains 5 balls, 3 new and 2 old. 2 balls are randomly selected one after another from the bag, the selected ball is not returned back. Find the probability that (a) 2 balls are new; (b) the 1 st ball is old one and the 2 nd ball is new; (c ) the 2 nd ball is the new one. 10/31/2020 Probability by Chtan -- FYHS Kulai 58

Soln : (a) Let A ={1 st ball is new ball}, B={2 nd ball is new ball} So, both are new balls=An. B 10/31/2020 Probability by Chtan -- FYHS Kulai 59

(b) 1 st ball is old ball=A’, P(A’)=2/5 2 nd ball is new ball ; P(B|A’)=3/4 1 st ball is old and 2 nd ball is new; P(A’n. B)=P(A’)x. P(B|A’) =(2/5) x (3/4) =3/10 10/31/2020 Probability by Chtan -- FYHS Kulai 60

(c) The event B is accompanied by A and A’, 2 possible cases. From (a) and (b), both An. B and A’n. B are mutually exclusive. Hence, P(B)= 3/10 + 3/10 = 3/5 10/31/2020 Probability by Chtan -- FYHS Kulai 61

We can use the tree diagram to do this e. g. 1 st time New 2 nd time Old New Old 10/31/2020 Probability by Chtan -- FYHS Kulai 62

e. g. 14 A ticket is to be given to 7 people. What is the chance for the 2 nd person to have the ticket? 10/31/2020 Probability by Chtan -- FYHS Kulai 63

Soln : Let A={1 st person get the ticket} B={2 nd person get the ticket} B is the dependent event of A. P(A)=1/7, then P(A’)=1 - 1/7 = 6/7 If event A occurs, the 2 nd person will not get the ticket! i. e. P(B|A)=0 10/31/2020 Probability by Chtan -- FYHS Kulai 64

If event A doesn’t occur, the 2 nd person will have a chance to get the ticket! P(B|A’)=1/6 10/31/2020 Probability by Chtan -- FYHS Kulai 65

1 st 1/7 6/7 2 nd person Get the ticket 0 1 1/6 No Get the ticket No 5/6 10/31/2020 Get the ticket Probability by Chtan -- FYHS Kulai No 66

Conclusion : The draw is fair to both persons, regardless of the 1 st or the 2 nd time to make selection. 10/31/2020 Probability by Chtan -- FYHS Kulai 67

Independent events 10/31/2020 Probability by Chtan -- FYHS Kulai 68

A and B are two events, the occurrence of A doesn’t affect the occurrence of B. They are called the independent events. 10/31/2020 Probability by Chtan -- FYHS Kulai 69

For example, throwing a die twice. The 1 st appearance of the outcome doesn’t influence the 2 nd throw. Another example, a ball is draw from a bag and put back and make a second draw. The outcomes are independent of each other. 10/31/2020 Probability by Chtan -- FYHS Kulai 70

We have the result: This is known as the multiplication law for independent events. 10/31/2020 Probability by Chtan -- FYHS Kulai 71

e. g. 15 Box A contains 6 white balls, 4 black balls and box B contains 3 white balls, 5 black balls. A ball is drawn from each box, find the probability that both balls are white? 10/31/2020 Probability by Chtan -- FYHS Kulai 72

Soln : Let A={white ball from box A}, B={white ball from box B} A and B are independent events. 10/31/2020 Probability by Chtan -- FYHS Kulai 73

We can use the tree diagram : Box B Box A 3/8 6/10 4/10 5/8 3/8 5/8 10/31/2020 Probability by Chtan -- FYHS Kulai 74

e. g. 16 In a shooting competition, both A and B have the same probability of 0. 6 to hit the target. Calculate the probability that (a) both hit the target ; (b) only one man hits the target; (c) at least one hits the target. 10/31/2020 Probability by Chtan -- FYHS Kulai 75

Soln : (a) Let A={A hits the target}, B={B hits the target} Both hit the target=An. B A and B are both independent events. =0. 6 x 0. 6 =0. 36 10/31/2020 Probability by Chtan -- FYHS Kulai 76

(b) Only one man hit the target=“A hits, B doesn’t hit” U “A doesn’t hit, B hits” 10/31/2020 Probability by Chtan -- FYHS Kulai 77

(c) Method 1 P(at least 1 hits the target)=P(only one hits)+P(both hit) =0. 36+0. 48=0. 84 Method 2 P(both miss)=P(A’n. B’)=P(A’)x. P(B’) =(1 - 0. 6)x(1 - 0. 6)=0. 16 P(at least 1 hits)=1 - 0. 16 = 0. 84 10/31/2020 Probability by Chtan -- FYHS Kulai 78

Summary Probability laws (1) For a finite probability space S with equally outcomes, and a subset E of S, 10/31/2020 Probability by Chtan -- FYHS Kulai 79

Probability by Chtan -- FYHS Kulai 10/31/2020 (2) If A and B are exhaustive, then If A and B are mutually exclusive, then Hence, Addition law for mutually exclusive events. 80

Probability by Chtan -- FYHS Kulai 10/31/2020 (3) So that, If A and B are independent, Multiplication law for independent events. 81

Probability by Chtan -- FYHS Kulai 10/31/2020 (4) If A and B are mutually exclusive, So that , 82

Probability by Chtan -- FYHS Kulai (5) 10/31/2020 If and then events A and B cannot be both independent and mutually exclusive, as 83

(6) or 10/31/2020 Probability by Chtan -- FYHS Kulai 84

Extension of results to more than two events 10/31/2020 Probability by Chtan -- FYHS Kulai 85

10/31/2020 Probability by Chtan -- FYHS Kulai 86

e. g. 17 In the Good Grub Restaurant customers may (if they wish) order any combination of chips, peas and salad to accompany the main course. The probability that a customer chooses salad is 0. 45, peas and chips 0. 19, salad and peas 0. 15, salad and chips 0. 25, salad or peas 0. 6, salad or chips 0. 84, salad or chips or peas 0. 9. Find the probability that a customer chooses (a) peas, (b) chips, (c) all three, (d) none of these. 10/31/2020 Probability by Chtan -- FYHS Kulai 87

Soln : Let A={salad}, E={peas}, C={chips} P(A)=0. 45, P(En. C)=0. 19, P(An. E)=0. 15, P(An. C)=0. 25, P(AUE)=0. 6, P(AUC)=0, 84, P(AUEUC)=0. 9 (a) P(peas are chosen)=P(E) P(AUE)=P(A)+P(E)-P(An. E) 0. 6=0. 45+P(E)-0. 15 P(E)=0. 3 10/31/2020 Probability by Chtan -- FYHS Kulai 88

(b) P(AUC)=P(A)+P(C)-P(An. C) 0. 84=0. 45+P(C)-0. 25 P(C)=0. 64 (c) P(AUEUC)=P(A)+P(E)+P(C)-P(An. C)P(An. E)-P(En. C)+P(An. En. C) 0. 9=0. 45+0. 3+0. 64 -0. 15 -0. 19 -0. 25+P(An. En. C)=0. 1 10/31/2020 Probability by Chtan -- FYHS Kulai 89

(d) P(customer chooses none)=P(A’n. E’n. C’) =1 – P(AUEUC) =1 – 0. 9 =0. 1 10/31/2020 Probability by Chtan -- FYHS Kulai 90

If events A, B and C are mutually exclusive, then P(AUBUC)=P(A)+P(B)+P(C) This can be extended to any number of mutually exclusive events : 10/31/2020 Probability by Chtan -- FYHS Kulai 91

The conditional probability P(An. B)=P(A). P(B|A) can be extended for 3 events A, B and C : P(An. Bn. C)=P[(An. B)n. C] =P(An. B). P[C|(An. B)] =P(A). P(B|A). P[C|(An. B)] 10/31/2020 Probability by Chtan -- FYHS Kulai 92

If A, B and C are independent events, then For n events, 10/31/2020 Probability by Chtan -- FYHS Kulai 93

Probability trees 10/31/2020 Probability by Chtan -- FYHS Kulai 94

A useful way of tackling many probability problems is to draw a ‘probability tree’. The method is illustrated in the following examples: 10/31/2020 Probability by Chtan -- FYHS Kulai 95

e. g. 18 A bag contains 8 white counters and 3 black counters. Two counters are drawn, one after the other. Find the probability of drawing one white and one black counter, in any order, (a) if the first counter is replaced, (b) if the first counter is not replaced. 10/31/2020 Probability by Chtan -- FYHS Kulai 96

Soln : (a) With replacement Let W 1 ={first counter is white} W 2={second counter is white} B 1={first counter is black} B 2={second counter is black} 10/31/2020 Probability by Chtan -- FYHS Kulai 97

1 1 / =8 ) 1 W P( P(B 1)= 3/1 1 1 st draw 10/31/2020 1 1 / 8 2)= P(W P(B 2)=3/11 1 1 / 8 2)= P(W P(B 2) =3/ 11 2 nd draw Probability by Chtan -- FYHS Kulai 98

P(drawing 1 white and 1 black)=P(W 1 n. B 2) + P(B 1 n. W 2) =24/121 + 24/121 = 48/121 10/31/2020 Probability by Chtan -- FYHS Kulai 99

(b) 10/31/2020 Without replacement Probability by Chtan -- FYHS Kulai 1 = l a Tot 100

P(drawing 1 white and 1 black)=P(W 1 n. B 2) + P(B 1 n. W 2) =24/110 + 24/110 = 48/110=24/55 10/31/2020 Probability by Chtan -- FYHS Kulai 101

e. g. 19 A fair coin is tossed three times. What is the probability of obtaining (a) exactly two heads, (b) at least two heads? 10/31/2020 Probability by Chtan -- FYHS Kulai 102

Soln : 2 / 1 = (H) (a) P 2 / 1 P (T)=1 = ) /2 (H P P(T )=1 /2 P(H) P(T =1/2 )=1/ 2 P(exactly 2 heads)=3/8 10/31/2020 2 / 1 = ) P(H P(T)=1/ 2 2 / 1 = ) H ( P P(T)=1/ 2 Probability by Chtan -- FYHS Kulai P(HHH)=1/8 P(HHT)=1/8 P(HTH)=1/8 P(HTT)=1/8 P(THH)=1/8 P(THT)=1/8 P(TTH)=1/8 P(TTT)=1/8 103

(b) P(at least 2 heads)=3/8 + 1/8 = 1/2 10/31/2020 Probability by Chtan -- FYHS Kulai 104

10/31/2020 Probability by Chtan -- FYHS Kulai 105

Mathematical Expectation 数学期望� 10/31/2020 Probability by Chtan -- FYHS Kulai 106

A man have a chance of p to receive x dollars, the product xp is called the mathematical expectation. In short, it is called Expectation. 期望值 For example, the probability of a boy receives 12 dollars is 1/6. Then, E=12 x (1/6)=2 10/31/2020 Probability by Chtan -- FYHS Kulai 107

e. g. 20 In a business activity, the businessman have a probability of 0. 6 to make a profit of 300 dollars. The probability that he will loss 100 dollars is 0. 4. Find the expectation value of this business. 10/31/2020 Probability by Chtan -- FYHS Kulai 108

Soln : Because we have 0. 6+0. 4=1, no other possibilities in this business. E = 300 x 0. 6 + (-100) x 0. 4 = 140 (dollars) This shows that he can hope to make profit in the business. 10/31/2020 Probability by Chtan -- FYHS Kulai 109

e. g. 21 The Magnum 4 -digit game has 10, 000 possible outcomes. The payouts for RM 1. 00 are : one 1 st prize RM 2, 000, one 2 nd prize RM 1, 000, one 3 rd prize RM 500, 10 special prizes RM 200 each and 10 consolation prizes RM 60 each. Calculate the mathematical expectation for paying RM 1. 00 to play the game. 10/31/2020 Probability by Chtan -- FYHS Kulai 110

Soln : 10/31/2020 Probability by Chtan -- FYHS Kulai 111

This is a losing game. It is worthless to play this game. 10/31/2020 Probability by Chtan -- FYHS Kulai 112

The Binomial Distribution 二�分配 10/31/2020 Probability by Chtan -- FYHS Kulai 113

Consider an experiment which has 2 possible outcomes, ‘success’ and ‘failure’. A binomial situation arises when n independent trials of the experiment are performed. 10/31/2020 Probability by Chtan -- FYHS Kulai 114

e. g. 22 A coin is biased so that the probability of obtaining a head is 2/3. The coin is tossed four times. Find the probability of obtaining exactly two heads. 10/31/2020 Probability by Chtan -- FYHS Kulai 115

Soln : Let H be the event ‘obtaining a head’ as success. Given P(H)=2/3, P(H’)=1/3 The probability of obtaining 2 tails and 2 heads, Independent events But the result ‘ 2 heads and 2 tails’ can be obtained in 10/31/2020 Probability by Chtan -- FYHS Kulai 116

Therefore , 10/31/2020 Probability by Chtan -- FYHS Kulai 117

e. g. 23 An ordinary die is thrown seven times. Find the probability of obtaining exactly three sixes. 10/31/2020 Probability by Chtan -- FYHS Kulai 118

Soln : We will consider ‘obtaining a 6’ as success. But ‘four not 6 and three 6’ can be obtained in ways 10/31/2020 Probability by Chtan -- FYHS Kulai 119

=0. 078 10/31/2020 Probability by Chtan -- FYHS Kulai 120

e. g. 24 The probability that a marksman hits a target is p and the probability that he misses is q, where q=1 -p. Write an expression for the probability that, in 10 shots, he hits the target 6 times. 10/31/2020 Probability by Chtan -- FYHS Kulai 121

Soln : P(success)=p, P(failure)=q=1 -p We require 4 failures and 6 successes, in any order, so 10/31/2020 Probability by Chtan -- FYHS Kulai 122

In general: If the probability that an experiment results in a successful outcome is p and the probability that the outcome is a failure is q, where q=1 -p, and if X is the r. v. (random variable). 10/31/2020 Probability by Chtan -- FYHS Kulai 123

‘the number of successful outcomes in n independent trials’, Then the p. d. f. of X is given by p. d. f : probability density function 10/31/2020 Probability by Chtan -- FYHS Kulai 124

Recall that a binomial expansion, 1= P(X=0) +P(X=1) +P(X=2) +…+P(X=r)+…+P(X=n) 10/31/2020 Probability by Chtan -- FYHS Kulai 125

If X is distributed in this way, we write where n is the number of independent trials and p is the probability of a successful outcome in one trial. n and p are called the parameters of the distribution. 10/31/2020 Probability by Chtan -- FYHS Kulai 126

e. g. 25 The probability that a person supports Party A is 0. 6. Find the probability that in a randomly selected sample of 8 voters there are (a) exactly 3 who support Party A, (b) more than 5 who support Party A. 10/31/2020 Probability by Chtan -- FYHS Kulai 127

Soln : Given p=0. 6 so q=1 -p=0. 4 Let X be the r. v. ‘the number of Party A supporters’ (a) 10/31/2020 =0. 124 Probability by Chtan -- FYHS Kulai 128

(b) =0. 315 10/31/2020 Probability by Chtan -- FYHS Kulai 129

e. g. 26 A box contains a large number of red and yellow balls in the ratio 1: 3. Balls are picked at random from the box. How many balls must be picked so that the probability that there is at least one red ball among them is greater than 0. 95? 10/31/2020 Probability by Chtan -- FYHS Kulai 130

Soln : Consider ‘obtaining a red ball’ as ‘success’. Then p=P(success)=1/4 and q=1 -p=3/4 Let X be the r. v. ‘the number of red ball’. Then X~Bin(n, p) where p=1/4 and n is unknown. Now, 10/31/2020 Probability by Chtan -- FYHS Kulai 131

We require Now So 10/31/2020 Probability by Chtan -- FYHS Kulai 132

i. e. The least value of n is 11. 10/31/2020 Probability by Chtan -- FYHS Kulai 133

Go home and do : 1. 2. Ans : 1(a) 0. 0823 (b) 0. 680 2(a) 0. 209 (b) 0. 0168 (c) 0. 00852 10/31/2020 Probability by Chtan -- FYHS Kulai 134

The Normal Distribution 常态分配 10/31/2020 Probability by Chtan -- FYHS Kulai 135

The normal distribution is the most important continuous distribution in statistics. Many measured quantities in the natural sciences follow a normal distribution, for example heights, masses, ages, random errors, I. Q. scores, examination results. 10/31/2020 Probability by Chtan -- FYHS Kulai 136

A continuous random variable X having p. d. f. f(x) where is said to have a normal distribution with mean and variance. 10/31/2020 Probability by Chtan -- FYHS Kulai 137

If X is distributed in this way we write f(x) x 10/31/2020 Probability by Chtan -- FYHS Kulai 138

There is a point of inflexion at And at. The actual size of the bell-shaped curve depends on the values of. 10/31/2020 Probability by Chtan -- FYHS Kulai 139

The distribution is bell shaped and symmetrical about x=. Approximately 95% of the distribution lies within Approximately 99. 8% of the distribution lies within 10/31/2020 Probability by Chtan -- FYHS Kulai 140

The range of the distribution is therefore approximately 6 standard deviations. The maximum value of f(x) occurs when and is given by 10/31/2020 Probability by Chtan -- FYHS Kulai 141

There is a point of inflexion at : The actual size of the bell-shaped curve depends on the value of and. 10/31/2020 Probability by Chtan -- FYHS Kulai 142

Here are some examples, each drawn to the scale : f -2 f 0. 4 0 0. 16 2 (1) X~N(0, 1) 10/31/2020 x 96 100 104 x (2) X~N(100, 6. 25) Probability by Chtan -- FYHS Kulai 143

f f 46 0. 2 50 x 52 (3) X~N(50, 4) 10/31/2020 0. 8 1 2 3 4 5 6 7 x (4) X~N(4, 1/4) Probability by Chtan -- FYHS Kulai 144

The red line is the standard normal distribution 10/31/2020 Probability by Chtan -- FYHS Kulai 145

The standard normal distribution 10/31/2020 Probability by Chtan -- FYHS Kulai 146

To standardise X, subtract divide by. So 10/31/2020 Probability by Chtan -- FYHS Kulai and 147

i. e. then 10/31/2020 Probability by Chtan -- FYHS Kulai 148

The probability density function for Z The p. d. f. of the standard normal variable Z is denoted by where 10/31/2020 Probability by Chtan -- FYHS Kulai 149

0. 4 -3 -2 -1 10/31/2020 0 1 2 Probability by Chtan -- FYHS Kulai 3 z 150

The cumulative distribution function for Z: The cumulative distribution function of the standard normal variable Z is denoted by where The integral is still very difficult to evaluate, so we refer to tables. 10/31/2020 Probability by Chtan -- FYHS Kulai 151

It should be noted that the tables may be printed in one of two different formats. (1) 0 10/31/2020 z Probability by Chtan -- FYHS Kulai 152

(2) 0 z The values of Q(z) are known as the ‘Uppertail Probabilities’. 10/31/2020 Probability by Chtan -- FYHS Kulai 153

Use of the standard normal tables using Only positive values of z are printed in the tables, so for negative values of z the symmetrical properties of the curve are used: P(Z<a)= a P(Z>-a)= -a P(Z>a)=1 a 10/31/2020 P(Z<-a)=1 -a Probability by Chtan -- FYHS Kulai 154

Note : 10/31/2020 We have Probability by Chtan -- FYHS Kulai 155

e. g. 27 If Z~N(0, 1), find from tables (a) P(Z<1. 377) (b) P(Z>-1. 377) (c) P(Z>1. 377) (d) P(Z<-1. 377) 10/31/2020 Probability by Chtan -- FYHS Kulai 156

Soln : (a) 0 1. 377 (b) -1. 377 0 (c) 0 1. 377 (d) 10/31/2020 -1. 377 0 Probability by Chtan -- FYHS Kulai 157

e. g. 28 If find (a) (b) (c) (d) (e) 10/31/2020 Probability by Chtan -- FYHS Kulai 158

Soln : 10/31/2020 Probability by Chtan -- FYHS Kulai 1. 751 0 0. 345 (a) 159

(b) -2. 696 10/31/2020 Probability by Chtan -- FYHS Kulai 0 1. 865 160

10/31/2020 -0. 6 -1. 4 (c) 0 Probability by Chtan -- FYHS Kulai 161

(d) -1. 433 10/31/2020 0 1. 433 Probability by Chtan -- FYHS Kulai 162

(e) -1. 527 10/31/2020 0 0. 863 Probability by Chtan -- FYHS Kulai 163

e. g. 29 The time taken by a milkman to deliver milk to the High Street is normally distributed with mean 12 minutes and standard deviation 2 minutes. He delivers milk every day. Estimate the number of days during the year when he takes (a) longer than 17 minutes, (b) less than 10 minutes, (c) between 9 and 13 minutes. 10/31/2020 Probability by Chtan -- FYHS Kulai 164

Soln : Let X be the r. v. ‘the time taken to deliver the milk to the High Street’. Then X~N(12, 4) We standardise X so that (a) s. d. =2 12 s. v. 0 10/31/2020 Probability by Chtan -- FYHS Kulai 17 2. 5 Standardised variable 165

Therefore The number of days when he takes longer than 17 minutes =365(0. 00621) =2. 27 Approximately 2 days in the year. 10/31/2020 Probability by Chtan -- FYHS Kulai 166

(b) s. d. =2 s. v. 10 -1 12 0 The number of days when he takes less than 10 minutes =365(0. 1587)=57. 9 58 10/31/2020 Probability by Chtan -- FYHS Kulai 167

(c) s. d. =2 s. v. 9 -1. 5 12 0 13 0. 5 The number of days=365(0. 6247)=228 days 10/31/2020 Probability by Chtan -- FYHS Kulai 168

The normal approximation to the binomial distribution 10/31/2020 Probability by Chtan -- FYHS Kulai 169

Under certain circumstances the normal distribution can be used as an approximation to the binomial distribution. One practical advantage is that calculations are much less tedious to perform. 10/31/2020 Probability by Chtan -- FYHS Kulai 170

If X~Bin(n, p) then E(X)=np Var(X)=npq Where q=1 -p Now, for large n and p not too small or too large, X ~ N(np, npq) approximately 10/31/2020 Probability by Chtan -- FYHS Kulai 171

e. g. 30 Find the probability of obtaining between 4 and 7 heads inclusive with 12 tosses of a fair coin, (a) using the binomial distribution, (b) using the normal approximation to the binomial distribution. 10/31/2020 Probability by Chtan -- FYHS Kulai 172

Soln : Let X be the r. v. ‘the number of heads obtained’. Let ‘success’ be ‘obtaining a head’. Then X ~ Bin(n, p) where n=12 and p=P(head)=1/2 (a) 10/31/2020 and Probability by Chtan -- FYHS Kulai 173

Now, =0. 121 So, 10/31/2020 Probability by Chtan -- FYHS Kulai 174

X~N(np, npq) where n=12 and p=1/2 So X~N(6, 3) Continuity corrections (b) Note : compares to (a), (b) is much quicker to perform! 10/31/2020 Probability by Chtan -- FYHS Kulai 175

Some examples of continuity corrections : 10/31/2020 Probability by Chtan -- FYHS Kulai 176

10/31/2020 Probability by Chtan -- FYHS Kulai 177

10/31/2020 Probability by Chtan -- FYHS Kulai 178

10/31/2020 Probability by Chtan -- FYHS Kulai 179

"The only real security that a man will have in this world is a reserve of knowledge, experience and ability. " -- Henry Ford, carmaker 10/31/2020 Probability by Chtan -- FYHS Kulai 180

"As long as you're going to be thinking anyway, think big. " -- Donald Trump, Real Estate Magnate 10/31/2020 Probability by Chtan -- FYHS Kulai 181

The 2009 year end examination scopes : 1. The straight line 2. The circle 3. The parabola 4. The ellipse 5. The hyperbola 排列与组合 6. Permutations & combinations 7. Probability概率 10/31/2020 hyperbola 182

nd 2 November, 2009 (Monday) S 2 S Mathematics Final Examination Part A : Short Questions -- answer all 9 questions x 5%=45% Part B : Long Questions -- answer 5 from 9 questions x 11%=55% 10/31/2020 hyperbola 183