Probability if `n` heads in `2n` tosses of a fair coin can be given by `prod_(r=1)^n((2r-1)/(2r))` b. `prod_(r=1)^n((n+r)/(2r))` c. `sum_(r=0)^n((^n C_r)/(2^n))` d. `(sumr=0n(^n C_r)^2)/(sumr=0 2n(^(2n)C_r)^)`

Options

A. `underset(r = 1)overset(n)prod((2r-1)/(2r))`
B. `underset(r = 1)overset(n)prod((n+r)/(2r))`
C. `underset(r = 0)overset(n)Sigma((.^(n)C_(r))/(2^(n)))^(2)`
D. `(underset(r = 0)overset(n)Sigma(.^(n)C_(r))^(2))/((underset(r=0)overset(2n)Sigma.^(2n)C_(r)))`

Answer

A, C, D

Explanation

`P(E) = (.^(2n)C_(n))/(2^(2n)) = ((2n)!)/(n! n! 2^(n) 2^(n))`
`=(1xx2xx3xx...xx(2n))/(n!n!2^(n)2^(n))`
`=(1xx3xx5...xx(2n - 1))/(n! 2^(n))`
Now,
`underset(r = 1)overset(n) prod((2r-1)/(2r))=(1xx3xx5xx...xx(2n-1))/(2xx4xx6xx...xx(2n))`
`=(1xx3xx5xx...xx(2n-1))/((1xx2xx3xx...xxn)2^(n))`
`underset(r = 0)overset(n)sum((.^(n)C_(r))/(2^(n)))^(2) = (1)/(2^(n)2^(n)) underset(r = 0)overset(n)sum (.^(n)C_(r))^(2) = (1)/(2^(n)2^(n)) .^(2n)C_(n)`
Also, `(underset(r = 0)overset(n)sum(.^(n)C_(r))^(2))/((underset(r = 0)overset(2n)sum .^(2n)C_(r))) = (.^(2n)C_(n))/(2^(2n))`