# Chapter 11 - 11.7 - Probability - 11.7 Exercises - Page 834: 20

$P(E)=\frac{1}{2}$

#### Work Step by Step

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: $E=[(H,H,H),(H,H,T),(H,T,H),(T,H,H)]$ $N(E)=4$ $P(E)=\frac{N(E)}{N(S)}=\frac{4}{8}=\frac{1}{2}$

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