$\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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