## Algebra and Trigonometry 10th Edition

$P(E)=\frac{1}{2}$
H -> head T -> tail All the possible outcomes (sample space): $S=[(H,H,H),(H,H,T),(H,T,H),(H,T,T),(T,H,H),(T,H,T),(T,T,H),(T,T,T)]$ We have that the total number of possible outcomes in the sample space is: $N(S)=8$ We want at least two heads (Event), that is, two heads or three heads: $E=[(H,H,H),(H,H,T),(H,T,H),(T,H,H)]$ $N(E)=4$ $P(E)=\frac{N(E)}{N(S)}=\frac{4}{8}=\frac{1}{2}$