InterviewSolution
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InterviewSolution
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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)^)`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)))` |
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Answer» Correct Answer - A::C::D `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))` |
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