$\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} $

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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 \(P(x_i),\: x_i = 0,1,2,3,4,5,6\)

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- 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 $\rightarrow P (X=3) = 20 \large(\frac{1}{2})^6$

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