Question

#9 (s) Casideracomtrwkichtte probability ofgeh heads 2a)Coisider a com tor Whi h heads and Nn tails 7

Answer 1

According to Binomial probability distribution, if a coin is randomly tossed N times, then, the probability of getting n heads is given by
$P(n)={^{N}\textrm{C}}_{n}p^nq^{N-n}$
where,
    ${^{N}\textrm{C}}_{n}=\frac{N!}{n!(N-n)!}$
And so, we have
$P(n)=\frac{N!}{n!(N-n)!}p^nq^{N-n}$

b)

Let say, if the coin is tossed N times, then, the number of heads is n. And so, if the gain value for head is a and lose value for tail is b, then, the net gain is
    $g=an-b(N-n)$
   $\Rightarrow g=(a+b)n-bN$
So, the expectation value of this is
   $\Rightarrow \langle g\rangle =\sum_{n=0}^N gP(n)=(a+b)\sum_{n=0}^NnP(n)-bN\sum_{n=0}^N P(n)$
where, P(n) is given by
   $P(n)=\frac{N!}{n!(N-n)!}p^nq^{N-n}$
So, we have
    $\sum_{n=0}^N P(n)=\sum_{n=0}^N\frac{N!}{n!(N-n)!}p^nq^{N-n}=(p+q)^N =1$
And
   $\sum_{n=0}^N nP(n)=\sum_{n=0}^N n \frac{N!}{n!(N-n)!}p^nq^{N-n}$
   $\Rightarrow \sum_{n=0}^N nP(n)=\sum_{n=1}^N \frac{N!}{(n-1)!(N-n)!}p^nq^{N-n}$
   $\Rightarrow \sum_{n=0}^N nP(n)=Np\sum_{n=1}^N \frac{(N-1)!}{(n-1)!(N-1-(n-1))!}p^{n-1}q^{N-1-(n-1)}$
    $\Rightarrow \sum_{n=0}^N nP(n)=Np\sum_{m=0}^{N-1} \frac{(N-1)!}{m!(N-1-m)!}p^{m}q^{N-1-m}$
where, n = m + 1 .
    $\Rightarrow \sum_{n=0}^N nP(n)=Np(p+q)^{N-1}=Np$
So, we get
$\Rightarrow \langle g\rangle =(a+b)Np-bN$
  average gain per toss is therefore
   $\frac{\langle g\rangle}{N} =(a+b)p-b$
   $\Rightarrow \frac{\langle g\rangle}{N} =ap-b(1-p)$
   $\Rightarrow \frac{\langle g\rangle}{N} =ap-bq$
As q = 1 - p.