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A coin is tossed 6 times. Find the probability of getting

(i) exactly 4 heads

(ii) at least 1 heads

(iii) at most 4 heads

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(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

\(=\frac{The\,favourable\,outcomes}{The\,total\,number\,of\, outcomes}\)

\(\Rightarrow 57/64\)

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