(i) As the coin is tossed 6 times the total number of outcomes will be \(2^6=64\)
And we know that the favourable outcomes of getting exactly 4 heads will be \(6_{C_4}=15\)
Thus, the probability of getting exactly 4 heads will be
\(=\frac{The\,favourable\,outcomes}{The\,total\,number\,of\,outcomes}\)
\(\Rightarrow 15/64\)
(ii) As the coin is tossed 6 times the total number of outcomes will be \(2^6=64\)
And we know that the favourable outcomes of getting at least 1 heads will be \(6_{C_1}+6_{C_2}+6_{C_3}+6_{C_4}+6_{C_5}+6_{C_6}=63\)
Thus, the probability of getting at least 1 head will be
\(=\frac{The\, favourable\, outcomes}{The\, total\, number\, of \,outcomes}\)
\(\Rightarrow 63/64\)
(iii) As the coin is tossed 6 times the total number of outcomes will be \(2^6=64\)
And we know that the favourable outcomes of getting at most 4 heads will be \(6_{C_0}+6_{C_1}+6_{C_2}+6_{C_3}+6_{C_4}=57\)
Thus, the probability of getting at most 4 heads will be
\(=\frac{The\,favourable\,outcomes}{The\,total\,number\,of\, outcomes}\)
\(\Rightarrow 57/64\)