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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 num

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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 means $n_H$ must be even. Possible values for $n_H$ are $0$ and $2$. Case $1$: $n_H = 0$ (all tails). The probability is $\binom{3}{0} (\frac{1}{2})^0 (\frac{1}{2})^3 = \frac{1}{8}$. Case $2$: $n_H = 2$ (two heads, one tail). The probability is $\binom{3}{2} (\frac{1}{2})^2 (\frac{1}{2})^1 = 3 	imes \frac{1}{8} = \frac{3}{8}$. The total probability is $\frac{1}{8} + \frac{3}{8} = \frac{4}{8} = \frac{1}{2}$.

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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