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

(C) $X$ is a random variable following a binomial distribution $B(6, 1/2)$.
Here,$n = 6$ and $p = 1/2$.
Therefore,$q = 1 - p = 1 - 1/2 = 1/2$.
The probability mass function is given by $P(X=x) = ^nC_x q^{n-x} p^x$.
Substituting the values,we get $P(X=x) = ^6C_x (1/2)^{6-x} (1/2)^x = ^6C_x (1/2)^6$.
Since $(1/2)^6$ is a constant,$P(X=x)$ is maximum when $^6C_x$ is maximum.
Calculating the values of $^6C_x$:
$^6C_0 = ^6C_6 = 6! / (0! 6!) = 1$
$^6C_1 = ^6C_5 = 6! / (1! 5!) = 6$
$^6C_2 = ^6C_4 = 6! / (2! 4!) = 15$
$^6C_3 = 6! / (3! 3!) = 20$
Comparing these values,$^6C_3 = 20$ is the maximum value.
Therefore,$P(X=3)$ is the maximum probability,which implies $X=3$ is the most likely outcome.