Question

Suppose X has a binomial distribution with n = 6 and \[p = \frac{1}{2} .\]  Show that X = 3 is the most likely outcome.

 

 

Answer 1

\[\text{ We have n = 6 and p }  = \frac{1}{2}\]
\[ \therefore q = 1 - p = \frac{1}{2}\]
\[\text{ Hence, the distribution is given by } \]
\[P(X = r) = ^{6}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{6 - r} , r = 0, 1, 2, 3, 4, 5, 6\]
\[ =^{6}{}{C}_r \left( \frac{1}{2} \right)^6 \]

\[P(X = r) = \frac{^{6}{}{C}_r}{2^6} \]
\[\text{ By substituting r = 0, 1, 2, 3, 4, 5 and 6, we get the following distribution for X }  . \]
    X      0    1    2   3     4    5    6
\[P(X) \frac{1}{64} \frac{6}{64} \frac{15}{64} \frac{20}{64} \frac{15}{64} \frac{6}{64} \frac{1}{64}\] 

Comparing the probabilities, we get that X = 3 is the most likely outcome.