Chapter 4 (continued) Nutan S. Mishra Department of Mathematics and Statistics University of South Alabama

Complementary events Given an event A, another event B in S is called its complementary if B contains all those points of S which are not in A. Consider the following pair of events in the experiment of rolling a die S={1,2,3,4,5,6} A = {2,4,6} B={1,3,5} Thus B contains all those points of S which are not in A. We call B as a complementary event of A. and denote it by A 2,4,

Complementary events In the experiment of rolling a die define A: occurrence of a number smaller than 3 A = {1,2} = {3,4,5,6} (non occurrence of A) P(A) = 2/6 then P( ) = 1-P(A) = 1-2/6 = 4/6 P( ) = 1-P(A)

Mutually exclusive events In a given sample space two events are mutually exclusive if they are disjoint That is no point is common between them. Another way to say : occurrence of one excludes the happening of the other. Example: consider rolling a die Define A: occurrence of an even number {2,4,6} Define B: occurrence if an odd number {1,3,5} 2,4,61,3,5 A B S

Mutually exclusive events Complementary events are always mutually exclusive because they do not have any sample point common between them. That is they can not occur together. Another Example of mutually exclusive events Gender of a newborn baby S {M, F} A: {F} B: {M} then A and B are mutually exclusive events.

Intersection of Two events Define event A: occurrence of a number smaller than 5 then A ={1,2,3,4} Event B: occurrence of number greater than 2, then B = {3,4,5,6} A  B is the event consists of the points which are in both A and B. Thus A  B = {3,4} A  B represents the event of occurrence of A and B together. 1,2 3, 4 5,6

Union of two events Define event A: occurrence of a number smaller than 5 then A ={1,2,3,4} Event B: occurrence of number greater than 2, then B = {3,4,5,6} AUB is the event consists of all those points which are either in A or in B AUB = {1,2,3,4,5,6} AUB represents the event where either A or B occurs. 1,2 3, 4 5,6

Independent events A pair of events in S say A and B are called independent of each other if either P(A|B) = P(A) or P(B|A) = P(B) In words: Probability of occurrence of A remains unaffected whether B occurs or not. Alternatively A and B are independent if occurrence of B remains same whether A has occurred or not.

Multiplication rule of probability P(A and B) is called joint probability of two events. P(A and B) = P(A  B) = P(A)*P(B|A) Alternatively P(A  B) = P(B)*P(A|B) If two events are independent then P(A  B) = P(A) * P(B). This is multiplication rule of probability. If two events are mutually exclusive then P(A  B) = 0

Exercise 4.70 Joint probability of A and B is (a) P(A  B) = P(A)*P(B|A) =.40*.25 =.1 (b) P(A  B) = P(B)*P(A\B) =.65*.36 = Given that two events are independent then their joint probability is given by P(A  B) = P(A)*P(B) (a) P(A  B) =.61*.27=.1647 (b) P(A  B) =.39*.63 =.2457

Exercise 4.75, (a) Joint probability of three events is given by P(A  B  C). Since all of them are independent P(A  B  C) = P(A)*P(B)*P(C) =.49*.67*.75=.2462 (b) P(A  B  C) =.71*.34*.45= P(B|A) = P(A  B)/P(A) =.24/.30 =.8

Exercise 4.80 AgeHave been victimized(V) Have never been victimized(NV) Marginal total 60-69(A) (B) r over(C) Marginal total (a) P(V  C) = 61/1800=.0339 This is the probability that a senior of age 80 or over from this group has victimized. P(NV  A) = 698/1800 =.3878 (b) P(B and C) = P(B  C) = 0/1800 = 0 This is the probability that a person belongs to B as well as C. This is impossible. Because B and C are mutually exclusive events.

Exercise 4.92 P1 =event that contractor wins first project P2=event that contractor wins second project P(P1) =P(P2) =.25 (a) P(P1  P2) =.25*.25 =.0625 (b) =.75*.75 = w L w L w L

Addition rule of probability (AUB) P(A or B) = P(AUB) = P(A)+P(B) - P(A  B) If the two events are independent then P(A  B)=P(A)*P(B) and then P(AUB) = P(A) + P(B) – P(A)*P(B) If the two events are mutually exclusive then P(A  B)=0 and then P(AUB) = P(A) + P(B)

Exercise AgeHave been victimized(V) Have never been victimized(NV) Marginal total 60-69(A) (B) r over(C) Marginal total P(V OR B) = p(V  B) = The selected person is either have been victimized or in the age group B P(VUB)= P(V) + P(B) – P(V  B) = 312/ / /1800 = 759/1800 =.4217 P(NV U C) = P(NV) + P(C) – P(NV  C) = 1488/ / /1800 =.8606