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(D) When three unbiased coins are tossed,the sample space $S$ is given by:$S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$The total number of outcomes

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(D) When three unbiased coins are tossed,the sample space $S$ is given by:$S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$The total number of outcomes is $n(S) = 8$.Let $E$ be the event of getting at most two heads.It is easier to calculate the complement event $E'$,which is the event of getting three heads.$E' = \{HHH\}$The number of outcomes in $E'$ is $n(E') = 1$.The probability of getting three heads is $P(E') = \frac{n(E')}{n(S)} = \frac{1}{8}$.The probability of getting at most two heads is $P(E) = 1 - P(E') = 1 - \frac{1}{8} = \frac{7}{8}$.

Three unbiased coins are tossed. Then,the probability of getting at most two heads is

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