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(A) Let $\lambda$ and $\mu$ be the number of heads obtained by $A$ and $B$ respectively,and $\lambda'$ and $\mu'$ be the number of tails obtained by $

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$1/2$

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(A) Let $\lambda$ and $\mu$ be the number of heads obtained by $A$ and $B$ respectively,and $\lambda'$ and $\mu'$ be the number of tails obtained by $A$ and $B$ respectively. Then $\lambda + \lambda' = n + 1$ and $\mu + \mu' = n$. We want to find the probability $P$ that $\lambda > \mu$. The probability of the complementary event $\lambda \le \mu$ is $1 - P$. Note that $\lambda \le \mu \iff n + 1 - \lambda' \le n - \mu' \iff 1 - \lambda' \le - \mu' \iff \lambda' \ge \mu' + 1$,which implies $\lambda' > \mu'$. Thus,$1 - P$ is the probability that $A$ gets more tails than $B$. By symmetry,the probability that $A$ gets more heads than $B$ is equal to the probability that $A$ gets more tails than $B$. Therefore,$P = 1 - P$,which gives $2P = 1$,or $P = 1/2$.

Two persons $A$ and $B$ have respectively $n + 1$ and $n$ coins which they toss simultaneously. Then the probability that $A$ will have more heads than $B$ is

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