A) \[\frac{(2n!)}{{{(n!)}^{2}}}{{\left( \frac{1}{2} \right)}^{2n}}\]
B) \[1-\frac{(2n!)}{{{(n!)}^{2}}}\]
C) \[1-\frac{(2n!)}{{{(n!)}^{2}}}\,.\,\frac{1}{{{4}^{n}}}\]
D) None of these
Correct Answer: C
Solution :
The required probability = 1 ? Probability of equal number of heads and tails \[=1-{{\,}^{2n}}{{C}_{n}}{{\left( \frac{1}{2} \right)}^{n}}{{\left( \frac{1}{2} \right)}^{2n-n}}=1-\frac{(2n)\,!}{n\,!\,n\,!}{{\left( \frac{1}{4} \right)}^{n}}=1-\frac{(2n)\,!}{{{(n\,!\,)}^{2}}}\cdot \frac{1}{{{4}^{n}}}\].
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