(i) As 10 coins are tossed simultaneously the total number of outcomes are \(2^{10}=1024.\)
The favourable outcomes of getting exactly 3 heads will be \(10_{C_3}=120\)
Thus, the probability
\(=\frac{The\,favourable\,outcomes}{The\,total\,number\,of\,outcomes}\)
\(=\frac{120}{1024}\)
\(=\frac{15}{128}\)
Hence, the probability is \(\frac{15}{128}.\)
(ii) As 10 coins are tossed simultaneously the total number of outcomes are \(2^{10}=1024.\)
The favourable outcomes of getting not more than 4 heads will be
\(10_{C_0}\) + \(10_{C_1}\) + \(10_{C_2}\) + \(10_{C_3}\) + \(10_{C_4}\) = 386
Thus, the probability
\(=\frac{The\,favourable\,outcomes}{The\,total\,number\,of\,outcomes}\)
\(=\frac{386}{1024}\)
\(=\frac{193}{512}\)
Hence, the probability is \(\frac{193}{512}.\)
(iii) As 10 coins are tossed simultaneously the total number of outcomes are \(2^{10}=1024.\)
The favourable outcomes of getting at least 4 heads will be
\(10_{C_4}\) + \(10_{C_5}\) + \(10_{C_6}\) + \(10_{C_7}\) + \(10_{C_8}\) + \(10_{C_9}\) + \(10_{C_{10}}\) = 848
Thus, the probability
\(=\frac{The\,favourable\,outcomes}{The\,total\,number\,of\,outcomes}\)
\(=\frac{848}{1024}\)
\(=\frac{53}{64}\)
Hence, the probability is \(\frac{53}{64}.\)