Given:
A coin is tossed 3 times.
To find the probability of getting at least two heads:
First, we write the sample space.
[tex]\begin{gathered} S=\mleft\lbrace\text{HHH, HHT, HTH, THH, HTT, TTH, THT, TTT}\mright\rbrace \\ n(S)=8 \end{gathered}[/tex]A be the event of getting at least two heads.
[tex]\begin{gathered} A=\mleft\lbrace\text{HHH, HHT, HTH, THH}\mright\rbrace \\ n(A)=4 \end{gathered}[/tex]So, the probability of getting at least two heads is
[tex]\begin{gathered} P(A)=\frac{n(A)}{n(S)} \\ =\frac{4}{8} \\ =\frac{1}{2} \end{gathered}[/tex]Hence, the probability is,
[tex]\frac{1}{2}[/tex]The event of getting at least two tails is also getting the same probability as at least two heads.
