Answer 1

When three coins are tossed once, the sample space is given by
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
i.e. n (S) = 8

Let E₂ = event of getting at least two heads
Then E₂ = {HHH, HHT, HTH, THH}
i.e. n (E₂) = 4

\[\therefore P\left( E_2 \right) = \frac{n\left( E_2 \right)}{n\left( S \right)} = \frac{4}{8} = \frac{1}{2}\]