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

Solution:

Given \mathrm{X} is any random variable whose binomial distribution is \mathrm{B}(6,1 / 2)

Thus, \mathrm{n}=6 and \mathrm{p}=1 / 2

q=1-p=1-1 / 2=1 / 2

Thus, P(X=x)={ }^{n} C_{x} q^{n-x} p^{x}, where x=0,1,2 \ldots n

={ }^{6} \mathrm{C}_{\mathrm{x}}\left(\frac{1}{2}\right)^{6-\mathrm{x}}\left(\frac{1}{2}\right)^{\mathrm{x}}

={ }^{6} \mathrm{C}_{\mathrm{x}}\left(\frac{1}{2}\right)^{6}

It can be clearly observed that \mathrm{P}(\mathrm{X}=\mathrm{x}) will be maximum if { }^{6} \mathrm{c}_{\mathrm{x}} will be maximum.

    \[\therefore{ }^{6} \mathrm{c}{\mathrm{x}}={ }^{6} \mathrm{C}{6}=1\]

    \[{ }^{6} \mathrm{c}{1}={ }^{6} \mathrm{c}{5}=6\]

    \[{ }^{6} \mathrm{c}{2}={ }^{6} \mathrm{C}{4}=15\]

    \[{ }^{6} \mathrm{C}_{3}=20\]

Hence we can clearly see that { }^{6} \mathrm{c}_{3} is maximum.

\therefore for x=3, P(X=x) is maximum.

Hence, proved that the most likely outcome is x=3.