Basics of Probability with Examples | Understanding Probability Theory

1. Chapter 16 Probability 概率 Probability by Chtan -- FYHS Kulai

2. 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 : Probability by Chtan -- FYHS Kulai

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

4. a-r n(S)=a S n(A)=r A Since, Probability by Chtan -- FYHS Kulai

5. 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. Probability by Chtan -- FYHS Kulai

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

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

8. or, Probability by Chtan -- FYHS Kulai

9. 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. Probability by Chtan -- FYHS Kulai

10. 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)=1-1/13=12/13 Probability by Chtan -- FYHS Kulai

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

12. 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, Probability by Chtan -- FYHS Kulai

13. With two dice By permutation, the possible outcomes of two dice is 6x6=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 Probability by Chtan -- FYHS Kulai

14. 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, P(A)=n(A)/n(S)=1/4 Probability by Chtan -- FYHS Kulai

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

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

17. The Venn diagram : S r-t s-t B A t Probability by Chtan -- FYHS Kulai

18. Probability by Chtan -- FYHS Kulai

19. 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? Probability by Chtan -- FYHS Kulai

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

21. Mutually exclusive events Probability by Chtan -- FYHS Kulai

22. 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. Probability by Chtan -- FYHS Kulai

23. In this case , When A and B are mutually exclusive events, and Probability by Chtan -- FYHS Kulai

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

25. 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. Probability by Chtan -- FYHS Kulai

26. 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. Probability by Chtan -- FYHS Kulai

27. 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. Probability by Chtan -- FYHS Kulai

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

29. 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. Probability by Chtan -- FYHS Kulai

30. 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 Probability by Chtan -- FYHS Kulai

31. (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 Probability by Chtan -- FYHS Kulai

32. 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 (b) 11/36 (c) 5/9 Probability by Chtan -- FYHS Kulai

33. 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),(6,6),(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 Probability by Chtan -- FYHS Kulai

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

35. Exhaustive events Probability by Chtan -- FYHS Kulai

36. 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. Probability by Chtan -- FYHS Kulai

37. 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. Probability by Chtan -- FYHS Kulai

38. (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. Probability by Chtan -- FYHS Kulai

39. 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. Probability by Chtan -- FYHS Kulai

40. 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(B)=1 – P(A) Probability by Chtan -- FYHS Kulai

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

42. "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 Probability by Chtan -- FYHS Kulai

43. Conditional Probability Probability by Chtan -- FYHS Kulai

44. 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) Probability by Chtan -- FYHS Kulai

45. Similarly, Probability by Chtan -- FYHS Kulai

46. Note : Probability by Chtan -- FYHS Kulai

47. Illustrating this by means of the Venn diagram, S n BA A B s-t AnB, t Probability by Chtan -- FYHS Kulai

48. This result is often written as : Probability by Chtan -- FYHS Kulai

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

50. 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. Probability by Chtan -- FYHS Kulai