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(A) Let $H$ denote a head and $T$ denote a tail. The points awarded are $2$ for $H$ and $1$ for $T$. When three coins are tossed,let $n_H$ be the numb

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$\frac{1}{2}$

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(A) Let $H$ denote a head and $T$ denote a tail. The points awarded are $2$ for $H$ and $1$ for $T$. When three coins are tossed,let $n_H$ be the number of heads and $n_T$ be the number of tails. The total points $S = 2n_H + 1n_T$. Since $n_H + n_T = 3$,we have $n_T = 3 - n_H$. Substituting this,$S = 2n_H + (3 - n_H) = n_H + 3$. For $S$ to be an odd number,$n_H + 3$ must be odd,which implies $n_H$ must be an even number. Possible values for $n_H$ are $0$ or $2$ (since $n_H \in \{0, 1, 2, 3\}$). Case $1$: $n_H = 0$ (all tails,$TTT$). Probability $P(n_H = 0) = \binom{3}{0} (\frac{1}{2})^0 (\frac{1}{2})^3 = \frac{1}{8}$. Case $2$: $n_H = 2$ (two heads,one tail,e.g.,$HHT, HTH, THH$). Probability $P(n_H = 2) = \binom{3}{2} (\frac{1}{2})^2 (\frac{1}{2})^1 = 3 	imes \frac{1}{8} = \frac{3}{8}$. The total probability of getting an odd number of points is $P(n_H = 0) + P(n_H = 2) = \frac{1}{8} + \frac{3}{8} = \frac{4}{8} = \frac{1}{2}$.

An unbiased coin is tossed to get $2$ points for turning up a head and $1$ point for the tail. If three unbiased coins are tossed simultaneously,then the probability of getting a total of an odd number of points is

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