Presentation on theme: "Chapter 4 Probability: Probabilities of Compound Events"— Presentation transcript:
1Chapter 4 Probability: Probabilities of Compound Events 4.1THE ADDITION RULE4.1.1 The General Addition Rule4.1.2The Special Addition Rule for Mutually Exclusive Events4.2 Conditional Probabilities4.3 The Multiplication Rule4.4 Independent Events and the Special Multiplication Rule4.4.1Independence of Two Events4.4.2Independence of More Than Two Events and the Special Multiplication Rule4.5 Bayes’ TheoremThe Total ProbabilityBayes’ Theorem
2P(AB) = P(A) + P(B) – P(AB) 4.1THE ADDITION RULE4.1.1 The General Addition RuleExample 1Events A and B are such that P(A) =19/30 , P(B) =2/5 and P(AB)=4/5 . Find P(AB).(Ans: 9/30)The general addition rule for two events, A and B, in the sample space S:P(AB) = P(A) + P(B) – P(AB)
3Example 2In 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 woman or someone who wears glasses? (Ans: 1/5)Example 3A class contains 10men and 20 women of which half the men and half the women have brown eyes. Find the probability p that a person chosen at random is a man or has brown eyes. (Ans: 2/3)
4P(ABC) = P(A) + P(B) + P(C) – P(AB) – P(AC) – P(BC) The General Addition Rule for Three EventsP(ABC) = P(A) + P(B) + P(C) – P(AB) – P(AC) – P(BC)+ P(ABC)
5P(A1A2…Ak) = P(A1) + P(A2) +… + P(Ak). 4.1.2 The Special Addition Rule for Mutually Exclusive EventsExample 1Records in a music shop are classed in the following sections:classical, popular, rock, folk and jazz. The respective probabilities that a customer buying a record will choose from each section are 0.3, 0.4, 0.2, 0.05 and Find the probability that a person (a) will choose a record from the classical or the folk or the jazz sections, (b) will not choose a record from the rock or folk or classical sections.If A1, A2, …, Ak are mutually exclusive, thenP(A1A2…Ak) = P(A1) + P(A2) +… + P(Ak).
64.2 Conditional Probabilities If A and B are two events and P(A) 0 and P(B) 0, then the probability of A, given that B has already occurred is written P(A|B) andP(A|B) =Example 1Given that a heart is picked at random from a pack of 52 playing cards, find the probability that it is a picture card.
7ExampleWhen a die is thrown, an odd number occurs. What is the probability that the number is prime?Two tetrahedral, with faces labelled 1,2,3 and 4, are thrown and the number on which each lands is noted. The ‘score’ is the sum of these two numbers. Find the probability that the score is even, given that at least one die lands on a 3.
84.3 The Multiplication Rule The general multiplication rule for events A and B in the sample space S:P(AB) = P(A) P(B|A)P(AB) = P(B) P(A|B)P(ABC) = P(A) P(B|A) P(C|AB)
94.4 Independent Events and the Special Multiplication Rule 4.4.1 Independence of Two EventsNote: If two evens are mutually exclusive, then P(AB) = _______. So for two events to be both independent and mutually exclusive we must have P(A) P(B) = P(AB) = ________. This is possible only if either P(A) = _________ or P(B) = __________.If the occurrence or non-occurrence of an event A does not influence in any way the probability of an event B, then event B is independent of event A and P(B|A) = P(B).Two events A and B are independent iff P(AB) = P(A)P(B)
10Example 1A die is thrown twice. Find the probability of obtaining a 4 on the first throw and an odd number on the second throw.Example 2A bag contains 5 red counters and 7 black counters. A counter is drawn from the bag, the colour is noted and the counter is replaced. A second counter is then drawn. Find the probability that the first counter is red and the second counter is black.Example 3A fair die is thrown twice. Find the probability that (a) neither throw results in a 4, (b) at least one throw results in a 4.Example 4Two events A and B are such that P(A) = , P(A|B) = and P(B|A) = .Are A and B independent events? (b) Are A and B mutually exclusive events?(c) Find P(AB). (d) Find P(B).
11P(A1 A2 …. Ak) = P(A1)P(A2)…P(Ak) Independence of More Than Two Events and the Special Multiplication RuleIf k events A1, A2,…., Ak are independent, thenP(A1 A2 …. Ak) = P(A1)P(A2)…P(Ak)
12Example 1A die is thrown four times. Find the probability that a 5 is obtained each time.Example 14Three men in an office decide to enter a marathon race. The respective probabilities that they will complete the marathon are 0.9, 0.7 and 0.6. Find the probability that at least two will complete the marathon. Assume that the performance of each is independent of the performances of the others.
13C.WConditional Probability1)In a family of two children with at least one girl. What is the probability that the other one is a boy?2)Suppose a box contains 3 white balls and 5 red balls.Balls are drawn randomly one by one without replacement from it. What is the probability that the second ball drawn will be red, given that the first ball drawn is white?Balls are drawn randomly one by one with replacement from it. What is the probability that the third ball drawn will be white, given that the first two balls drawn are white.
143)A credit card company has surveyed new accounts from university students. Suppose a samples of 160 students indicated the following information in terms of whether the student possessed a credit card X and/or a credit card Y.credit card Xcredit card YYesNo50203060
154.) Let event A = students possessed two credit cards. event B = students possessed at least one credit card.event C = students did not possess any card.event D = students possessed a credit card X.event E = students possessed a credit card Y.Find the probabilities of each of these events A,B,C,D,E,Find also and .. Find also,.
165) A fair coin is tossed three times Let event A = Head appears on first toss.event B = Head appears on second toss.event C = Head appears on all three tosses.To find whether A and B, B and C, C and A are independent.
17P(A) = P(E1)P(A|E1) + P(E2)P(A|E2) +…+ P(Ek)P(A|Ek) 4.5 Bayes’ TheoremThe Total ProbabilitySuppose a sample space S is partitioned into k mutually exclusive events Ej (j = 1,2,…,k), i.e. S = E1E2….Ek with EiEj = for ij, thenP(A) = P(E1)P(A|E1) + P(E2)P(A|E2) +…+ P(Ek)P(A|Ek)=
184.5.2 Bayes’ TheoremLet the sample space S be partitioned into mutually exclusive events Ej’s (j = 1,2,…,k) and let A be an event in S. Then the probability of Er conditional on A isP(Er |A) = for r =1,2,…,k
19Suppose there are three identical boxes which contain different number of white and black balls. A box is selected at random and a ball is drawn from it randomly .(I) What is the probability that a white ball is chosen?(ii) Suppose a white ball is chosen, find the probability that this white ball comes from the 1st box.Number of white ballsNumber of black balls1 st box832 nd box653 rd box47
202) The marketing manager of a soft drink manufacturing firm is planning to introduce a new rand of Coke into the market. In the past, 30 % of the Coke introduced by the company have been successful, and 70% have not been successful. Before the Coke is actually marketed, market research is conducted and a report, either favorable or unfavourable, is compiled. In the past, 80% of the successful Coke received favourable reports and 40% of the unsuccessful Coke also received favourable reports. The marketing manager would like to know the probability that the new brand of Coke will be successful if it receives a favourable report.
213)A man decided to visit his friend at North Point. He can reach there by MTR, Bus or Tram respectively. The following information is given:(i) He was late for his visit. Find the probability that he had travelled by MTR.(ii) He was not late for his visit. Find the probability that he had travelled by Bus.Probability of being takenProbability of being lateMTR5/81/4Bus2/85/9Tram1/87/8