Classical Definition of Probability Relative Frequency Definition of Probability Some properties of Probability

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1 PROBABILITY Recall that in a random experiment, the occurrence of an outcome has a chance factor and cannot be predicted with certainty. Since an event is a collection of outcomes, its occurrence cannot be predicted with certainty either (except for impossible and the certain events). However, it is often desirable to know how likely an event will occur. It is common practice to assign, estimate or derive from some basic postulates the chance with which an event will occur. This gives an incentive to the study of probability. The probability of an event may be defined in several ways. We shall consider only the classical only the classical definition and the relative frequency definition. Classical Definition of Probability Let S be an equiprobable sample space with n(s) sample points and A be an event in S with n(a) sample points. Then the probability of A, denoted by P(A), is P(A) = n(a) n(s) Relative Frequency Definition of Probability Suppose that a random experiment is repeated a large number of times N, and that the event A occurs n times. Then the probability of A is the limiting value of the relative frequency n N as N becomes infinitely large. Some properties of Probability From the definition of probability(whichever definition is used), we may derive the following properties: For every event A in the sample space S, 1. 0 < P(A) < 1 2. P(S) =1 3. If A and B are mutually exclusive events in S, then P(A U B) = P(A) +P(B) :P(impossible event) =0 P(the certain event)= 1

2 Law for complementary events A useful application of the above properties gives the probability law for complementary events A and A in the sample space S: P(A ) = 1-P(A) IN a soccer match between two team A and B, the probability that team A will win is 1/4 and the probability that tam B will win is 1/3. Find the probability that (a) team A or team B will win the match, (b) the two team tie. SOLUTION: (a).p(team A or team B wins)=p(a)+p(b) = 1/4+ 1/3 = 7/12 (b).p(the two teams tie) = 1- P(team A or team B wins) =1-7/12 =5/12 We will also need to use the method of counting in probability: SUCH AS Permutations & COMBINATION. The addition rule: Two events may occur individually or together. The probability that either one or the other or both will occur is given by the following rule. General addition rule: (a) For two events P(A B) P(A) P(B) P(A B) A card is drawn randomly from the desk of 52 cards, what is the probability that it is an ACE or Spade? Solution: P( the card drawn is an ACE)=4/52 P(the card drawn is a Spade)=13/52 P(ACE AND SPADE)=4+13/52-1/52=4/52 (b) For three events P(A B C) P(A) P(B) P(C) P(A B) P(B C) P(A C) P(A B

3 (A1 A2 A3 An) P(A1) P(A2) P(A3)+ +P(An) If both the events are mutually exclusive. Let A and B be two events, such that P(A) > 0, the probability of B given that A has Occurred can be found by: P( B A) = Two fair dice are rolled. Find the probability that (a) both numbers turned up are the same, given that the sum of the number is greater than 6. (b) The sum of the numbers turned up is greater than 6, given that both numbers are the same. Solution: (a) Apply the formula above, P(the sum of the number is greater than 6)= (3/36) / (21/36/) =1/7 (b) (c) P(both numbers are the same) =(3/36) / (6/36) =1/2 The multiplication rule 1. For two events P(A B) P(A) P(B A) or P(B) P(A B) 2. For three events P(A B C) P(A) P(B A) P(C A B) THIS RULE CAN BE EXPERSSED BY TREE DIAGRAM. THE WORK PROCEDURE CAN BE FOUND IN THE BOOK.

4 Independent events and the special multiplication rule For k events P(A1 A2 Ak) P(A1)P(A2) P(Ak) i.e. P(B A) P(B), the probability. of B occurring is not affected by the occurrence of A. It is known that 20% of the pipes produced by a manufacturer are substandard. If 3 of the pipers are selected at random, what is the probability that (a): all are substandard? (b):none is substandard? (c): exactly 1 is substandard? SOLUTION: (a): P(all are substandard)=0.2x0.2x0.2=0.008 (b): P(None is substandard)=0.8x0.8x0.8=0.512 (c): P(exactly 1 is substandard):3x0.2x0.8=0.384 Bayes Theorem 1. The total Probability Rule If an event A must result in one of the mutually exclusive events E1, E2,, En, then P(A) P(E1) P(A E1) P(E2) P(A E2) P(En) P(A En) 2. Bayes Theorem (Reversal of conditioning) (a) For two events P( E1 A)=P(E1)P(A E1)/ P(E2)P(A E2) EXERCISE 1. (a ) Write down the sample space of the sex patterns of the children of a 2-child family in the order of their ages. (You may use B to denote a boy and G to denote a girl.) (b) Assume that having a boy or having a girl is equally likely. It is known that a family has two children and they are not both girls. (i) Write down the sample space of the sex patterns of the children in the order of their ages.

5 (ii) What is the probability that the family has two sons? 2. In asking some sensitive questions such as Are you homosexual?, a randomized response technique can be applied : The interviewee will be asked to draw a card at random from a box with one red card and two black cards and then consider the statement I am homosexual if the card is red and the statement I am not homosexual otherwise. He will give the response either True or False. The colour of the card drawn is only known to the interviewee so that nobody knows which statement he has responded to. Suppose in a survey, 790 out of 1200 interviewees give the response True. (a) Estimate the percentage of persons who are homosexual. (b) For an interviewee who answered True, what is the probability that he is really homosexual? 3. A and B are two independent events. If P(A) = 0.4 and P(A B) = 0.7, find P(B). 4. In the election of the Legislative Council, 48% of the voters support Party A, 39% Party B and 13% Party C. Suppose on the polling day, 65%, 58% and 50% of the supporting voters of Parties A, B and C respectively cast their votes. (a) A voter votes on the polling day. Find the probability that the voter supports Party B. (b) Find the probability that exactly 2 out of 5 voting voters support Party B. 5. A flower shop has 13 roses of which 2 are red, 5 are white and 6 are yellow. Mary selects 3 roses randomly and the colours are recorded. (a) Denote the red rose selected by R, the white rose by W and the yellow rose by Y. List the sample space (i.e. the set of all possible combinations of the colours of roses selected, for example, 1R 2W denotes that 1 red rose and 2 white roses are selected). (b) Find the probability that Mary selects exactly one red rose. (c) Given that Mary has selected exactly one red rose, find the probability that only one of the other two roses is white.

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