Question

A fair coin is tossed $2n$ times, then the probability that the outcomes do not result in an equal number of heads and tails is

• A. $1-\frac{(2n)!}{(n!)^{2}}(\frac{1}{2})^{2n}$
• B. $1-\frac{(2n)!}{(n!)^{2}}$
• C. $\frac{(2n)!}{(n!)^{2}}(\frac{1}{2})^{2n}$
• D. $\frac{(2n)!}{(n!)^{2}}$

Hint:

In $2n$ tosses, the "central" term of the binomial expansion is always the most likely single outcome, but as $n$ increases, its probability $\frac{\binom{2n}{n}}{2^{2n}}$ actually approaches zero (approx. $\frac{1}{\sqrt{n\pi}}$).

Answer 1

The Correct Option is A

We need to find the probability that the number of heads is NOT equal to the number of tails in $2n$ tosses. It is easier to find the probability of the complementary event (equal heads and tails) and subtract it from 1.

Step 1: \color{redDefine the Complementary Event
Total tosses $N = 2n$.
For heads and tails to be equal, there must be exactly $n$ heads and $n$ tails.

Step 2: \color{redCalculate the Probability of Equal Heads and Tails
This is a Binomial distribution problem with $N = 2n$, $k = n$, and $p = 1/2$.
$P(\text{Heads} = n) = \binom{2n}{n} p^n (1-p)^{2n-n}$
$P(\text{Heads} = n) = \binom{2n}{n} (1/2)^n (1/2)^n$
$P(\text{Heads} = n) = \binom{2n}{n} (1/2)^{2n}$

Step 3: \color{redExpand the Binomial Coefficient
$\binom{2n}{n} = \frac{(2n)!}{n!(2n-n)!} = \frac{(2n)!}{(n!)^2}$.
So, $P(\text{Equal}) = \frac{(2n)!}{(n!)^2} \left(\frac{1}{2}\right)^{2n}$.

Step 4: \color{redFind the Required Probability
The probability of "not equal" is $1 - P(\text{Equal})$:
$P(\text{Not Equal}) = 1 - \frac{(2n)!}{(n!)^2} \left(\frac{1}{2}\right)^{2n}$.
This corresponds to Option (1).