An unbiased coin is tossed thrice. The probab | vedclass

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(A) The total number of possible outcomes when a coin is tossed thrice is $2^3 = 8$. These outcomes are: $HHH, HHT, HTH, THH, HTT, THT, TTH, TTT$. We

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$\frac{1}{2}$

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(A) The total number of possible outcomes when a coin is tossed thrice is $2^3 = 8$. These outcomes are: $HHH, HHT, HTH, THH, HTT, THT, TTH, TTT$. We are looking for the event where the number of heads is greater than the number of tails. The favorable outcomes are: $1$. $HHH$ ($3$ heads,$0$ tails) $2$. $HHT$ ($2$ heads,$1$ tail) $3$. $HTH$ ($2$ heads,$1$ tail) $4$. $THH$ ($2$ heads,$1$ tail) Thus,the number of favorable outcomes is $m = 4$. The required probability is $\frac{m}{n} = \frac{4}{8} = \frac{1}{2}$.

An unbiased coin is tossed thrice. The probability of receiving more heads than tails is ........

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