since(TH), (TT)are no longer possible. Secondly, the a-field taken to become FA={SA,,{(HT)},{(HH)}} Thirdly the probability set function become PA(SA)=1,P4(∞)=0,PA({(HT)})=,P4({(HH)} Thus, knowing that event A-one h has occurred in the first trial transformed the original probability space(S, F, p) to the conditional probability space (SA, FA, PA). The question that naturally arises is to what extent we can de- rive the above conditional probabilities without having to transform the original probability space. The following formula provides us with a way to calculate the conditional probability P(41)=P(A14)=P(41n4 xaMl Let A1=I(HT)) and A=((HT), (HH)J, then since P(A1)=4, P(a)=3 P(A1∩4)=P({(HT)})=士 /41 Using the above rule of conditional probability we can deduce that P(A1∩A2)=P(A1|42)P(A2) P(A2A1)·P(A1)forA1,A2∈F This is called the multiplication rule. Moreover, when knowing that A2 has occurred does not change the original probability of A1, i.e P(A1|A2)=P(A1) we say that Al and A2 are independentsince (T H),(TT) are no longer possible. Secondly, the σ-field taken to become FA = {SA, ∅, {(HT)}, {(HH)}}. Thirdly the probability set function become PA(SA) = 1, PA(∅) = 0, PA({(HT)}) = 1 2 , PA({(HH)} = 1 2 . Thus, knowing that event A-one H has occurred in the first trial transformed the original probability space (S, F,P) to the conditional probability space (SA, FA,PA). The question that naturally arises is to what extent we can de￾rive the above conditional probabilities without having to transform the original probability space. The following formula provides us with a way to calculate the conditional probability. PA(A1) = P(A1|A) = P(A1 ∩ A) P(A) . Example: Let A1 = {(HT)} and A = {(HT),(HH)}, then since P(A1) = 1 4 , P(A) = 1 2 , P(A1 ∩ A) = P({(HT)}) = 1 4 , PA(A1) = P(A1|A) = 1/4 1/2 = 1 2 , as above. Using the above rule of conditional probability we can deduce that P(A1 ∩ A2) = P(A1|A2) · P(A2) = P(A2|A1) · P(A1) for A1, A2 ∈ F. This is called the multiplication rule. Moreover, when knowing that A2 has occurred does not change the original probability of A1, i.e. P(A1|A2) = P(A1), we say that A1 and A2 are independent. 11