Probability of odd-man when $n$ people tossing a coin with probability $p$ of getting head Οdd-man means a person gets a different result from all other people. Probability of getting a head is $p$. Number of coins knowing the number of people is $n$. So number of possible ways to get someone different of all the others are  $\dbinom{n}{1} = n $. For me the result is $np^{n-1}q^1$, while the book says $$npq(p^{n-2}+q^{n-2})$$
 A: There are two ways in which one person gets a result that differs from everybody else.


*

*1 person tosses a heads and everyone else gets a tails.

*1 person tosses a tails and everyone else gets a heads.


The probability of the first event is $nq^{n-1}p$. We derive this term by observing that there are $n$ ways of choosing the odd one out (since there are $n$ people and like you pointed out this is the same as $\binom{n}{1}$) and the probability of each such event is $q^{n-1}p$.
Similarly, for the second event we get $np^{n-1}q$.
Since either of these two cases may occur, we add these two probabilites and get the final answer as
$$npq(p^{n-2} + q^{n-2})$$
A: An odd-man can occur in two ways: Either $n-1$ persons toss heads and the odd-man tosses tails or $n-1$ persons toss tails and the odd-man tosses heads. With $q:=1-p$


*

*probability of $n-1$ heads and $1$ tail: $\displaystyle\dbinom{n}{1}p^1q^{n-1}$

*probability of $n-1$ tails and $1$ head: $\displaystyle\dbinom{n}{1}p^{n-1}q^{1}$


Hence, summing up (or of events translates to $+$ of probabilities) gives the probability of odd-man $$\dbinom{n}{1}p^1q^{n-1}+\dbinom{n}{1}p^{n-1}q^1=npq(p^{n-2}+q^{n-2})$$
