Suppose X has a binomial distribution $$B\: ( 6, \frac{1}{2})$$ . Show that X = 3 is the most likely outcome. Hint : P(X = 3) is the maximum among all $$P(x_i),\: x_i = 0,1,2,3,4,5,6$$

Toolbox:
• For any Binomial distribution $B (n, p),$ the probability of x success in n-Bernoulli trials, $P (X = x) = \large^{n}C_x. p^x.q^{n–x}$ where $x = 0, 1, 2,...,n$ and $(q = 1 – p)$
For any Binomial distribution $B (n, p),$ the probability of x success in n-Bernoulli trials, $P (X = x) = \large^{n}C_x. p^x.q^{n–x}$ where $x = 0, 1, 2,...,n$ and $(q = 1 – p)$
Given a binomial distribution $B\; \large($$6, \large\frac{1}{2})$$ \rightarrow n = 6$ and $p = \large \frac{1}{2} \rightarrow$ $q = \large\frac{1}{2}$
$P (X = x) = \large^{6}C_x. \large\frac{1}{2}^x.\frac{1}{2}^{6–x} = \large^{6}C_x.\large(\frac{1}{2})^6$
We can see that the the most likely outcome is when $\large^{6}C_x$ is maximum.
We can calculate $\large^{6}C_x$ for $x = 0,1,2,3...6$ to see which yields the maximum value:
$\begin{matrix} x & 0 &1 &2 &3 &4 &5&6 \\ ^{6}C_x& 1&6 &15 &20 &15 &6 &1 \end{matrix}$
The most likely outcome is the outcome whose probability is the highest.
Therefore, $X=3$ has the maximum of all above values $\rightarrow P (X=3) = 20 \large(\frac{1}{2})^6$