Explanation:
A) 1){H, H}
2){H, T}
3){T, H}
4){T, T}
B) The coin is balanced, so the probability is the same for Tail or head, the first throw is independent of the second. then I can multiply their probabilities:
P(H1∩H2)=P(H)*P(H)=1/2*1/2=1/4
P(H1∩T2)=P(H)*P(H)=1/2*1/2=1/4
P(T1∩H2)=P(H)*P(H)=1/2*1/2=1/4
P(T1∩T2)=P(H)*P(H)=1/2*1/2=1/4
C) A={{T, H}, {H, T}}, B={{H, H}, {H, T}, {T, H}}
D) P(A)=P(H1∩T2)+P(T1∩H2)=1/4+1/4=1/2
P(B)=P(H1∩H2)+P(H1∩T2)+P(T1∩H2)=1/4+1/4+1/4=3/4
P(A∩B)=P(H1∩T2)+P(T1∩H2)=P(A)=1/2
P(AUB)=P(H1∩H2)+P(H1∩T2)+P(T1∩H2)=P(B)=3/4
Ac={{T, T}, {H, H}}
P(AcUB)=P(H1∩H2)=1/4