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10 coins are tossed simultaneously. Find the probability of getting

(i) exactly 3 heads

(ii) not more than 4 heads

(iii) at least 4 heads

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(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 \(\frac{53}{64}.\)

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