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Basics of Probability with Examples | Understanding Probability Theory

Learn about the fundamentals of probability theory, including definitions, calculations, and practical examples to enhance your understanding. Discover how to calculate probabilities for different scenarios with clear explanations.

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Basics of Probability with Examples | Understanding Probability Theory

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  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

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