Question-12
If a fair coin is tossed an even number \(2n\) times, then what is the probability of obtaining more heads than tails?
Start with defining random variables as \(X\) represents number of heads and \(Y\) represents number of tails. Given, total number of trials is \(2n\) and consider \(P(X>Y) + P(X<Y) + P(X=Y) = 1\)
Let \(X\) represents number of heads and \(Y\) represents number of tails. Total number of trials is \(2n\).
\(P(X>Y) + P(X<Y) + P(X=Y) = 1\hspace{20mm} \ldots (i)\)
Since, by symmetry, \(p=1-p=\dfrac{1}{2} \implies P(X>Y) = P(X<Y)\). Therefore, equation \((i)\) will be: \(2P(X>Y) = 1 - P(X = Y)\hspace{20mm} \ldots (ii)\)
In \(2n\) trials, \(X = Y \implies\) Number of heads \(=\) Number of tails \(=n\). Thus, the equation \((ii)\) will be: \(2 P (X > Y ) = 1 - ^{2n}C_{n}\left(\dfrac{1}{2}\right)^{n}\left(\dfrac{1}{2}\right)^{n}\)
\(\implies P(X > Y) = \dfrac{1}{2}\left\{1-^{2n}C_{n}\left(\dfrac{1}{2}\right)^{2n}\right\}\)