Suppose X has a binomial distribution B (6,1 2 ). Show that X = 3… - Vidyadip

Question

Suppose X has a binomial distribution $\text{B}\Big(6,\frac{1}{2}\Big).$ 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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Answer

X is the random variable whose binomial distribution is $\text{B}\Big(6,\frac{1}{2}\Big).$
Therefore, n = 6 and $\text{p}=\frac{1}{2}$
$\therefore\ \text{q}=1-\text{p}=1-\frac{1}{2}=\frac{1}{2}$
$\text{Then}\ \text{P}(\text{X=x})=\ ^\text{n}\text{C}_\text{x}\text{q}^\text{n-x}\text{p}^\text{x}$
$=\ ^6\text{C}_\text{x}\bigg(\frac{1}{2}\bigg)^{6-\text{x}}.\bigg(\frac{1}{2}\bigg)^\text{x}$
$=\ ^6\text{C}_\text{x}\bigg(\frac{1}{2}\bigg)^{6}$
It can be seen that P(X = x) will be maximum, if $ ^6\text{C}_\text{x}$ will be maximum.
$\text{Then},\ ^6\text{C}_{0}=\ ^6\text{C}_{6}=\frac{6!}{0!\cdot6!}=1$
$\ ^6\text{C}_{1}=\ ^6\text{C}_{5}=\frac{6!}{1!\cdot5!}=6$.
$\ ^6\text{C}_{2}=\ ^6\text{C}_{4}=\frac{6!}{2!\cdot4!}=15$
$\ ^6\text{C}_{3}=\frac{6!}{3!\cdot3!}=20$
The value of $ ^6\text{C}_{3}$ is maximum. Therefore, for x = 3, P(X = x) is maximum.
Thus, X = 3 is the most likely outcome.

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