Q.4.15 - A First Course in Probability

Question

Suppose that n independent tosses of a coin having probability p of coming up heads are made. Show that the probability that an even number of heads results is 

$\frac{1}{2}\left[1+(q-p{)}^n\right]$.where$q=1-p$.Do this by proving and then utilizing the identity

$\sum _{i=0}^{[n/2]}\left(\begin{array}{c}n\\ 2i\end{array}\right)p^{2i}{q}^{n-2i}=\frac{1}{2}\left[(p+q{)}^n+(q-p{)}^n\right]$

Step-by-Step Solution

Answer

In the given information the answer is $=\frac{1}{2}\left[1+(q-p{)}^n\right]$ isproved

Step 1: Given Information

We know that ,

${\left(p+q\right)}^n$$=\sum _{i=0}^n{}^{n}C_k{p}^k{q}^{n+1}$(1)

${\left(q-p\right)}_n$ $(q-p{)}^n=\sum _{i=0}^n{}^{n}C_k(-p{)}^k{q}^{n-k}$ (2)

Step 2 : Calculation

When we add these two expressions ,

$\Rightarrow {\left(p+q\right)}^n+{\left(q-p\right)}^n=$$2\sum _{i=0}^{\frac{n}{2}}{C}_{2i}{p}^{2i}{q}^{n-2i}$

because when $k$ is odd ${\left(-p\right)}^k$ is negative so  (1) and (2) will cancel .

 $\Rightarrow \sum _{i=0}^{\frac{n}{2}}{\mathrm{C}}_{2\mathrm{i}}{\mathrm{p}}^{2\mathrm{i}}{\mathrm{q}}^{\mathrm{n}-2\mathrm{i}}=\frac{1}{2}\left[(p+q{)}^n+(q-p{)}^n\right]$

$\Rightarrow p\left(Even heads\right)$=$\frac{1}{2}\left[(p+1-p{)}^n+(q-p{)}^n\right]$

$=\frac{1}{2}\left[1+(q-p{)}^n\right]$ hence proved

Step 3 : Final Answer

The answer is $=\frac{1}{2}\left[1+(q-p{)}^n\right]$ is proved$\frac{1}{2}\left[1+(q-p{)}^n\right]$