Question

Probability of $n$ heads in $2n$ tosses of a fair coin can be given by

• A. $\prod _{r=1}^n\left(\frac{2r-1}{2r}\right)$
• B. $\prod _{r=1}^n\left(\frac{n+r}{2r}\right)$
• C. $\sum _{r=0}^n{\left(\frac{{}^n{C}_r}{2^n}\right)}^2$
• D. $\frac{\sum _{r=0}^n{(}^n{C}_r{)}^2}{\left(\sum _{r=0}^{2n}{}^{2n}{C}_r\right)}$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $\prod _{r=1}^n\left(\frac{2r-1}{2r}\right)$
C $\sum _{r=0}^n{\left(\frac{{}^n{C}_r}{2^n}\right)}^2$
D $\frac{\sum _{r=0}^n{(}^n{C}_r{)}^2}{\left(\sum _{r=0}^{2n}{}^{2n}{C}_r\right)}$

Probability of $n$ heads in $2n$ tosses of a fair coin $={}^{2n}{C}_n\times {\left(\frac{1}{2}\right)}^n\times {\left(\frac{1}{2}\right)}^n$

$\to {}^{2n}{C}_n\times {\left(\frac{1}{2}\right)}^n\times {\left(\frac{1}{2}\right)}^n=\sum _{r=0}^n{\left(\frac{{}^n{C}_r}{2^n}\right)}^2$

$\to {}^{2n}{C}_n\times \frac{1}{2^n}\times \frac{1}{2^n}=\frac{2n!}{n!\times n!\times 2^{2n}}=\frac{2^n.n!\times \mathrm{1.3.5}\dots \dots (2n-1)}{n!\times n!\times 2^{2n}}$

$=\frac{\mathrm{1.3.5}\dots \dots (2n-1)}{(n!)\times \left(2^n\right)}=\prod _{r=1}^n\left(\frac{2r-1}{2r}\right)$


$\to \prod _{r=1}^n\left(\frac{n+r}{2r}\right)=\frac{(n+1)(n+2)\dots \dots 2n}{\mathrm{2.4.6}\dots \dots 2n}$

$=\frac{2n!}{n!\cdot 2^n\cdot n!}=\frac{{}^{2n}{C}_n}{2^n}$


$\frac{\sum _{r=0}^n{(}^n{C}_r{)}^2}{\left(\sum _{r=0}^{2n}{}^{2n}{C}_r\right)}=\frac{({}^n{C}_0{)}^2+\dots \dots \dots +({}^n{C}_n{)}^2}{{}^{2n}{C}_0+\dots \dots \dots +{}^{2n}{C}_{2n}}=\frac{{}^{2n}{C}_n}{2^{2n}}$