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प्रश्न
Suppose X has a binomial distribution with n = 6 and \[p = \frac{1}{2} .\] Show that X = 3 is the most likely outcome.
उत्तर
\[\text{ We have n = 6 and p } = \frac{1}{2}\]
\[ \therefore q = 1 - p = \frac{1}{2}\]
\[\text{ Hence, the distribution is given by } \]
\[P(X = r) = ^{6}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{6 - r} , r = 0, 1, 2, 3, 4, 5, 6\]
\[ =^{6}{}{C}_r \left( \frac{1}{2} \right)^6 \]
\[P(X = r) = \frac{^{6}{}{C}_r}{2^6} \]
\[\text{ By substituting r = 0, 1, 2, 3, 4, 5 and 6, we get the following distribution for X } . \]
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.
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