As the coin is tossed 6 times the total number of outcomes will be 26
And we know that the favourable outcomes of getting at least 3 heads will be \(6_{C_3}+6_{C_4}+6_{C_5}+6_{C_6}\)
Thus, the probability of getting at least 3 heads will be
\(=\frac{The\,favourable\, outcomes}{The\,total\,number\,of\,outcomes}\)
\(\Rightarrow=\frac{(\frac{6}{3})+(\frac{6}{4})+(\frac{6}{5})+(\frac{6}{6})}{2^6}\)
\(\Rightarrow=\frac{21}{32}\)
