# Suppose X has a binomial distribution $B\; \large($$6, \large\frac{1}{2}). What is the most likely outcome and for what value of X, where x = 0, 1, 2, 3, 4, 5, 6? \begin{array}{1 1} X = 3, \; P (X=3) = \large\frac{20}{64} \\ X = 3, \; P (X=4) = \large\frac{15}{64} \\ X = 3, \; P (X=4) = \large\frac{25}{64} \\ X = 3, \; P (X=3) = \large\frac{15}{64} \end{array} ## 1 Answer Toolbox: • For any Binomial distribution B (n, p), the probability of x success in n-Bernoulli trials, P (X = x) = \large^{n}C_x. p^x.q^{n–x} where x = 0, 1, 2,...,n and (q = 1 – p) For any Binomial distribution B (n, p), the probability of x success in n-Bernoulli trials, P (X = x) = \large^{n}C_x. p^x.q^{n–x} where x = 0, 1, 2,...,n and (q = 1 – p) Given a binomial distribution B\; \large($$6, \large\frac{1}{2})$$\rightarrow n = 6$ and $p = \large \frac{1}{2} \rightarrow$ $q = \large\frac{1}{2}$
$P (X = x) = \large^{6}C_x. \large\frac{1}{2}^x.\frac{1}{2}^{6–x} = \large^{6}C_x.\large(\frac{1}{2})^6$
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 $\rightarrow P (X=3) = 20 \large(\frac{1}{2})^6$