Question

In Problem 7.70, suppose that the coin is tossed $n$times. Let $X$denote the number of heads that occur. Show that $P\{X=i\}=\frac{1}{n+1}i=0,1,\dots ,n$

Hint: Make use of the fact that,

${\int }_0^1{x}^{a-1}(1-x{)}^{b-1}dx=\frac{(a-1)!(b-1)!}{(a+b-1)!}$

When $a$and$b$ are positive integers.

Short answer

It has been shown that$P\{X=i\}=\frac{1}{n+1};i=0,1,\dots ,n$

Step-by-step solution

Given Information

Number of times coins tossed $=n$

Number of heads occur$=X$

Use:${\int }_0^1{x}^{a-1}(1-x{)}^{b-1}dx=\frac{(a-1)!(b-1)!}{(a+b-1)!}$

Positive integers$=a,b$

Explanation

$X=$Number of heads that occur

If coin is tossed $n$times then $X$can take value $0.1.2\dots \dots ,n$

$P[Headsoccur]=p~U(0,1)$

$\mathrm{P}[\mathrm{X}=\mathrm{i}]={\int }_0^1\left(\begin{array}{l}\mathrm{n}\\ \mathrm{i}\end{array}\right){\mathrm{p}}_i(1-\mathrm{p}{)}^{n-i}\mathrm{dp};i=0,1,2,\dots ,n$

$=\frac{\mathrm{n}!}{\left(\mathrm{i}\right)!(\mathrm{n}-\mathrm{i})!}{\int }_0^1 {\mathrm{p}}^{\mathrm{i}}(1-\mathrm{p}{)}^{\mathrm{n}-\mathrm{i}}\mathrm{dp}$

$=\frac{\mathrm{n}!}{\left(\mathrm{i}\right)!(\mathrm{n}-\mathrm{i})!}\frac{\left(\mathrm{i}\right)!(\mathrm{n}-\mathrm{i})!}{(\mathrm{n}+1)!}$

Since

${\int }_0^1{x}^{a-1}(1-x{)}^{b-1}dx=\frac{(a-1)!(b-1)!}{(a+b-1)!}$

$\therefore \mathrm{P}[\mathrm{X}=\mathrm{i}]=\frac{1}{n+1};i=0,1,2,\dots ,n$

Final Answer

Hence, it has been shown that$P\{X=i\}=\frac{1}{n+1};i=0,1,\dots ,n$