A fair coin is tossed, and a fair die is thrown. Write down sample spaces for
(a) the toss of the coin;{H,T}, P(H)=P(T)=1/2(b) the throw of the die;{1,2,3,4,5,6}, P(1)=P(2)=P(3)=P(4)=P(5)=P(6)=1/6(c) the combination of these experiments.{H&1,H&2,H&3,H&4,H&5,H&6,T&1,T&2,T&3,T&4,T&5,T&6}
P(H&1)=P(H&2)=P(H&3)=P(H&4)=P(H&5)=P(H&6)=
P(T&1)=P(T&2)=P(T&3)=P(T&4)=P(T&5)=P(T&6)=(1/2)(1/6)=1/12Let A be the event that a head is tossed,A = {H}and B be the event that an odd number is thrown.B = {1,3,5}Directly from the sample space, calculate P(A ∩ B)P(A ∩ B) = P(A&B) = P({H}&{1,2,3}) = P[(H&1)&(H&3)&(H&5)] = (1/12)(1/12)(1/12) = (1/12)^3 = 1/1728and P(A ∪ B)P(A ∪ B) = P(A and/or B) = P({H} and/or {1,2,3}) =
[That's all the ones with either an [H or a (1 or 2 or 3)] or both,
an [H and a (1 or 2 or 3)]
P({H&1,H&2,H&3,H&4,H&5,H&6,T&1,T&2,T&3}) = P(H&1)+P(H&2)+P(H&3)+P(H&4)+
P(H&5)+P(H&6)+P(T&1)+P(T&2)+P(T&3) =
(1/12)+(1/12)+(1/12)+(1/12)+(1/12)+(1/12)+(1/12)+(1/12)+(1/12) =
(1/12)∙9 = 9/12 = 3/4
Edwin